Grade 9 (Alternate) Mathematics III - Learning Modules for EASE Program of DepEd

Compilation by Ben: r_borres@yahoo.com

COMPILATION
OF LEARNING MODULES

GRADE 9
(Alternative)

MATHEMATICS III

Effective and Alternative
Secondary Education
(EASE)
Third Year

Compilation by Ben: r_borres@yahoo.com
CONTENTS

01.   Module 1 ‐ Circles
02.   Module 2 ‐ Circles
03.   Module 1 Properties of Quadrilaterals
03.   Module 2 ‐ Triangle Trigonometry
04.   Module 2 Properties of Quadrilaterals
05.   Module 1 ‐ Geometric Relations
08.   Module 1 Similarity
09.   Module 2 ‐ Similarity
10.   Module 3 ‐ Similarity
11.   Module 1 ‐ Plane Coordinate Geometry
12.   Module  2 ‐ Plane Coordinate Geometry
13.   Module 3 Plane Coordinate Geometry
14.   Module 1‐ Triangle Congruence
16.   Module 3 ‐ Triangle Congruence
17.   Module 1 Geometry of Shape and Size
18.   Module 2 ‐ Geometry of Shape and Size
20.   Module 4‐ Geometry of Shape and Size

Module 1
Circles
What this module is about
This module will discuss in detail the characteristics of a circle as well as the
segments and lines associated with it. Here, you will gain deeper understanding of the
angles formed in circles, how to get their measures and how they are related to one
another. Furthermore, this module will also give meaning to the circle being composed of
arcs and how each arc is related to the angles formed in circles.
What you are expected to learn
This module is written for you to
1. define a circle.
2. define and show examples of the lines and segments associated with circles.
3. describe the relationship of lines and segments that are peculiar to circles.
4. define, identify and give examples of the kinds of arcs that compose a circle.
5. identify central angle and inscribed angle.
6. discover the relationship between the measures of central angle and inscribed angle
and their intercepted arcs.
How much do you know
Answer the following as indicated. S
1. Given a circle with center O. Name the following :
a. the circle T
b. a diameter M
c. two radii OO
d. two chords which are not diameters N
e. a secant
f. a tangent R
2. If a radius is perpendicular to a chord then it ________ the chord.
·

2
3. In the given circle A, PT is a diameter, therefore
MT is a ________ and
4. PTM is a _________.
5. Radius AB ⊥ CE. If CE = 8 cm, then
CX = ________.
6. Using the same figure, if AX = 3 cm,
What is the length of radius AC?
7. In circle O, 93=∠BOCm . What is mBC?
8. What is BACm∠ ?
9. In the figure, PR ║ ST . Using the given
Find mPT and RPSm∠ .
10. A quadrilateral PQRS is inscribed in a circle.
If 103=∠Pm , what is Rm∠ ?
What you will do
Lesson 1
Identifying a circle, the lines, segments and angles associated with it.
A circle is defined as the set of all points that are at the same distance from a given
point in the plane. The fixed given point is called the center. The circle is named after its
center. Hence in the figure, given is a circle O. A
The set of points on the plane containing the circle is
divided into 3, (1)
the circle, (2)
the set of points outside
the circle and (3)
the set of points inside the circle.
OC , OB and OA are segment whose endpoints are
the center of the circle and a point on the circle.
These three segments are called radii of the circle.
Radius of a circle is a segment whose endpoints are the center and a point on the
circle. In the figure, AD is a segment whose endpoints are points on the circle. AD is
called chord of the circle. AB is a segment whose endpoints are points on the circle and it
D
B
C
O
O
B
A
C
P
M●
●
A
T
A
X
E
B
C
P R
ST

3
passes through the center. AB is called diameter of a circle. Diameter of a circle is a chord
that passes through the center.
Lines on the plane containing the circle may intersect the circle at one point or at two
points or not at all.
Fig. 1. line a does not Fig. 2. line b intersect Fig. 3 line c intersect circle
intersect circle O. circle O at point X at two points R and S.
In figure 2, line b is tangent to the circle, and in figure 3, line c is a secant. Hence, we can
use the following definitions.
Tangent is a line that intersect a circle at one points. Secant is a line that intersect a circle
at two points.
Some theorems in circle show relationship between chord and radius. One of them
is this theorem:
Theorem: If a radius is perpendicular to a chord, then it bisects the chord.
Proof: Consider the given circle. If radius OA ⊥ BC at D, then OA
bisects BC or BD = DC. One way of proving segments or angles
congruent is by showing that they are corresponding parts of
congruent triangles. Here, we must prove that BD and DC are
corresponding sides of congruent triangles. If O and B are
joined and O and C are also joined, we have ∆OBD and ∆OCD.
Both of these triangles are right since BCOA ⊥ and thus ODB∠ and
ODC∠ are both right angles. Since OB and OC are both radii of the same circle,
hence they are congruent. And finally OBOD ≅ by reflexive property. Therefore, by
the HyL Congruency for right triangles, OCDOBD ∆≅∆ . Since the two triangles are
congruent, then the remaining corresponding parts such as BD and DC are also
congruent.
We have just proven the theorem here, only this time, instead of using the two
column form we use the paragraph form.
a
O
X
O O
b
c
S
R
B
A
C
o
D
···
·

4
Our conclusion therefore is that a radius that is perpendicular to a chord bisects the
chord. The most important considerations here were the perpendicularity and the word to
bisect.
Examples:
1. DEOB ⊥ at T, DT = 3x -7 , TE = x + 15
Solution:
Since ⊥OB DE , then DT = TE
Hence,
3x – 7 = x + 15
2x = 15 + 7
2x = 22
x = 11
Substituting the value of x, we get
DT = 3(11) – 7 = 33 – 7
= 26
TE = 11 + 15 = 26
DE = DT + TE
DE = 26 + 26 = 52
There are other theorems whose main idea is taken from the previously proven
theorem. The next theorem serves as the converse of the first theorem and it states that: If
a radius of a circle bisects a chord that is not a diameter, then it is perpendicular to the
chord.
If the previous theorem was proven using the HyL congruence for right triangle, the
converse is proven using the reverse process, that is two angles must be proven part of
congruent triangles and they are congruent and supplementary.
You can prove the theorem as part of your exercise. Examples on how to use these
two theorems are given below.
2. Given: AB bisects chord CD at E.
CD = 6, AE = 4
Find the length of the radius of the circle.
Solution: Based on the theorem, CDAB ⊥ , thus
ADEACE ∆≅∆ and both are right triangles.
By the Pythagorean theorem, we can solve
for the length of the radius.
In ∆ACE, AC 2
= AE 2
+ CE 2
But CE = ½ CD so
T
A
EC D
B
O
D E
B
T

5
CE = ½ (6)
CE = 3
AC 2
= AE 2
+ CE 2
AC 2
= 42
+ 32
AC 2
= 16 + 9 = 25
AC = 25
AC = 5
Lesson on circle is very rich with theorems and definitions, principles and postulates.
Some of those theorems and definitions will be introduced as we plod along with this
module.
Definitions:
• Congruent circles are circles that have congruent radii.
• Concentric circles are coplanar circles having the same center
Illustrations:
a) b)
Circle A is congruent to circle B if and These two circles are
only if BYAX ≅ concentric circles
Theorem:
If chords of a circle or of congruent circles are equidistant from the center(s), then the
chords are congruent
Illustration of the theorem.
Circle O ≅ circle P
OX = PY
Then, CDAB ≅
A
X
B
Y
O
X
A B
O
X
N
P
C D
Y
M
·

6
Try this out
A. Using the given figure, name
1. the circle
2. 2 diameters
3. 2 chords which are not diameters
4. 2 secants
5. a tangent
B. Given: CDAB ⊥ at E
CD is 10 cm long. How far is CD
from the center if the length of the
radius is
1. 13 cm 5. 12 cm
2. 7 cm 6. 10 cm
3. 14 cm 7. 5 2 cm
4. 8 cm 8. 63 cm
C. Given: CD is 20 cm long. How long is the radius of the
circle if the distance of CD from the center is
1. 7 cm 3. 13 cm
2. 10 cm 4. 8 cm
5. 5 cm 7. 55 cm
6. 21 cm 8. 64 cm
D. AC is 12 cm long. How long is chord CD if its distance from the center is
1. 10 cm 5. 9 cm
2. 6 cm 6. 23 cm
3. 8 cm 7. 112 cm
4. 5 cm 8. 54 cm
E. Solve the following problems.
1. MPON ⊥
ME = 7x + 5
PE = 4x – 20
Solve for ME , PE and MP
O
M PE
N
A
C D
B
E
A
B
D
E
C
F
O

7
2. In a circle are two chords whose lengths are 10 cm and 24 cm respectively. If the
radius of the circle is 13 cm, what is the maximum distance of the two chords?
What is their minimum distance?
Lesson 2
Arcs and Central Angles
A part of a circle between any two points is an arc. In the
figure, the set of points from A to B is an arc. A circle is in itself
an arc. Arc of a circle is measured in terms of degrees.
The whole arc making up the circle measures 360°.
Any arc of a circle can belong to any of these three groups.
a. minor arc – an arc whose measure is between 0 and 180°.
b. semicircle – an arc whose measure is exactly 180°
c. major arc – an arc whose measure is between 180° and 360°
In the given figure, AB is a diameter, hence AB represents
a semicircle, AC is minor arc and ABC is a major arc.
Aside from AC, another minor arc in the figure is BC.
ACB also represents a semicircle.
Angles in a circle are formed by radii, chords, secants and
tangents. Determination of the measures of the angles formed by
these lines depends upon the measure of the intercepted arcs
of the given angles.
Examples:
In circle some angles formed by chords and
radii are shown. Each of the angles intercepts
an arc defined by the endpoints contained on the
sides of the angle.
∠ AEB intercepts AB.
∠ BOC intercepts BC
∠ COD intercepts CD
∠ EOD intercepts ED
O
A
B
C
D
E
A
C
B
O●
A
O
B• •
·

8
B
83°
D
At this point we will discuss in detail the kinds of angles formed in a circle, their
characteristics and how to get their measures from the measures of the intercepted arcs.
We will start with the angle formed by two radii.
Central angle is an angle formed by two radii
and the vertex is the center of the circle. In the
figure, ∠ AOB, ∠ BOD and ∠ DOC are all
examples of central angles. Each of these angles
has its own intercepted arc. ∠ AOB intercepts AB,
∠ BOD intercepts BD and ∠ DOC intercepts DC.
The measure of a central angle is numerically
equal to its intercepted arc.
In the figure, ∠ BAC is a central angle and
∠ BAC intercepts BC. Since mBC = 83, then
m∠ BAC = 83, mBDC = 277°.
In the study of geometry, every new topic or concept is always associated with study
of postulates, theorems and definitions. In the study of arcs and angles in a circle, we will
discuss many theorems that will help us solve problems involving the said concepts. We
will start with the simplest postulate in the chapter.
Like any measure, measure of an arc is also a unique real number and as such, we
can perform the four fundamental operations on those measure. So the first postulate is the
Arc Addition Postulate: The measure of an arc formed by two adjacent non-overlapping
arcs is the sum of the measures of the two arcs.
In the given circle, m AC = m AB + m BC
Examples:
1. DG is a diameter. Find the measure of the
following arcs.
DG, DE, DF, GE, DGF
Solution:
Since DG is a diameter, then DG is a semicircle.
Therefore,
m DG = 180
m DE = 180 – (60 + 70)
= 180 – 130
= 50
O
A
B
D
C
A
C
•
●
•
•
A
B
C
F
·
·
·
·
●
60°
70°
D
E
G
O
·

9
60°
60°
60°
A
B
C
D
m DF = m DE + m EF
= 50 + 60
= 110
m GE = m GF + m FE
= 70 + 60
= 130
m DGF = m DG + m GF
= 180 + 70
= 250
Definitions:
In the same circle or in congruent circles, arcs which have the same measure are
congruent.
Example: 1. In the figure, m DC = 60, m BC = 60
m AB = 60 .
Therefore, DC≅BC ≅ AB
2. Since every semicircle measures 180°,
then all semicircles are congruent.
Theorem:
If two minor arcs of a circle or of congruent circles are congruent, then the
corresponding chords are congruent.
Examples: 1. Given: AB ≅ BC
Since AB subtends AB and
BC subtends BC then
AB ≅ BC
2. Circle O ≅ circle M
If AB ≅ XY, then XYAB ≅
Theorem:
If two chords of a circle or of congruent circles are congruent, then the corresponding
minor arcs are congruent.
This is the converse of the previous theorem. Basically if you prove these two
theorems, the steps will be just the reverse of the other. Instead of proving them, showing
examples will be more beneficial to you.
●
●
O •
A
B
M •
X
Y
A
B
C·

10
In circle A, if PQRS ≅
then RS ≅ PQ
Theorem:
If two central angles of a circle or of congruent circles are congruent, then the
corresponding minor arcs are congruent.
Example: In circle O, ∠ MNO≅ ∠ BOA
Therefore, MP ≅ AB
Theorem:
If two minor arcs of a circle or of congruent circles are congruent, then the
corresponding central angles are congruent.
Example:
In circle A, DEBC ≅
Therefore `DAEBAC ∠≅∠
Theorem:
If two central angles of a circle or of congruent circles are congruent, then the
corresponding chords are congruent.
Given: In circle O, AOBXOY ∠≅∠
Prove: XY AB≅
Proof
Statements Reasons
1. In circle O, AOBXOY ∠≅∠
2. OBOX ≅ , OAOY ≅
3. BOAXOY ∆≅∆
4. XY AB≅
1. Given
2. Radii of the same or congruent circles are
congruent
3. SAS congruency Postulate
4. Corresponding parts of congruent
triangles are congruent..
A
B
C
D
E
X
B
Y
A
O
Q
A •
R
P
S
O
A
B
P
M

11
Theorem:
If two chords of a circle or of congruent circles are congruent circles are congruent,
then the corresponding central angles are congruent.
Given: In circle A, STPR ≅
Prove: SATPAR ∠≅∠
Proof:
Statements Reasons
1. In circle A, STPR ≅
2. ASAP ≅
ATAR ≅
3. SATPAR ∆≅∆
4. SATPAR ∠≅∠
1. Given
2. Radii of the same circle are
congruent.
3. SSS Congruency Postulate
4. Corresponding parts of congruent
triangles are congruent
Examples:
Given: AB and CD are diameters of circle E.
1. What is true about AED∠ and BEC∠ ? Why?
2. What kind of angles are they?
3. Give as many conclusions as you can
based on the previously discussed theorems.
Answers:
1. BECAED ∠≅∠ . They are vertical angles and vertical angles are congruent.
2. In the circle they are central angles. Central angles are angles whose vertex is the
center of the circle.
3. a. AD ≅ BC. If two central angles of a circle or of congruent circles are congruent,
then the corresponding arcs are congruent.
b. BCAD ≅
Likewise
1. BEDAEC ∠≅∠
2. AC ≅ DB
3. DBAC ≅
Try this out
A. AB is a diameter of circle O.
82=∠AOEm .
A
D
B
C
E
A
p
R
S
T

12
Find the measures of:
1. AB 4. ABE
2. AE 5. BAE
3. BE
B. GE and FD are diameters of circle A. If DA = 73º, find the measures of
1. ∠ DAE 5. GF
2. ∠ GAF 6. DG
3. ∠ EAF 7. FDE
4. ∠DAG
C. Given circle A. If m BTY = 116,
and m = 3n, find
1. m 5. m ∠BAT
2. n 6. m ∠TAY
3. BT 7. m ∠BAY
4. TY
D. Given circle O. AB ≅ BC. If mAB = 56, what is
AOBm∠ ? What is mABC ?
Which chords are congruent?
E. A. B and C are three points on the circle.
IF AC ≅ AB ≅ BC, what is the measure of each arc?
What is true about the chords ,, ABAC and BC ?
If ABC is 16 more than three times AC,
find mAC, mABC.
F. P, Q and R are three points on a circle. If the
ratio PQ:QR:PR = 3:4:5, find the measures of
PR, QR and PS.
O
A
E
B
A
D
E
G
F
O
A
B
C
A
B T
Y
n m
·
A
C
B

13
G. Using the figure and the given in it, find the
measures of:
1. PQ 5. POQ∠
2. QR 6. ∠QOR
3. SR 7. ∠SOR
4. PS 8. ∠POS
H. BD and EC are diameters of circle A.
If 35=∠Cm , find the measures of
1. ∠B 5. ∠EAD
2. ∠E 6. BC
3. ∠D 7. CD
4. ∠BAC
Lesson 3
Arcs and Inscribed Angles
Another angle in a circle that is very important in the study of circle is the inscribed
angle.
Definition:
An inscribed angle is an angle whose vertex lies on the circle and the sides contain
chords of the circle.
Fig. 1 Fig. 2 Fig 3
Each of the angle shown above is an example of an inscribed angle. Three cases
are represented here relative to the position of the sides in relation to the center of the
circle.
A
B
E
C
D
35º
A
B
C
●
P
S
T●
D
E
F
●
O
x
R
S
2x + 13
2x 4x + 5
P
Q

14
Case 1. the center of the circle is on one side of the inscribed angle.
Case 2, the center of the circle is in the interior of the inscribed angle.
Case 3, the center of the circle is on the exterior of the inscribed angle.
In the study of the angles in a circle and in determining their measures, it is important
to determine the intercepted arc(s) of the given angle. To understand better, let us see
some examples.
In the figure, the arc in the interior
of the angle is the intercepted arc of
the angle.
The intercepted arc of BAC∠ is
the minor arc AC.
In the given examples of inscribed angles above the following holds:
a) In figure 1, ∠ DEF is an inscribed angle
∠ DEF intercepts arc DF
b) In figure 2, ∠ PST is an inscribed angle,
∠ PST intercepts arc PT
c) In figure 3, ∠ BAC is an inscribed angle
∠ BAC intercepts arc BC
Every angle whether in a circle on in any plane is associated with a unique number
defined as its measure. If the measure of a central angle is equal to the measure of its
intercepted arc, the next theorem will tell us how to find the measure of the inscribed angle.
Theorem: Inscribed angle Theorem
The measure of an inscribed angle is equal to one half the measure of its intercepted
arc.
It means that in the given figure,
mDFDEFm 2
1
=∠
Intercepted arc
B A
C
D
E
F
O ●
●

15
Since there are three cases by which an inscribed angle can be drawn in a circle, then we
have to prove each of those cases.
Case 1 (One side of the angle is the diameter of the circle)
Given: Circle O with inscribed angle DEF∠
Use the notation in the figure for
clarity
Prove: ( )mDFDEFm
2
1
=∠
Proof:
Statements Reasons
1. Circle O with inscribed angle DEF∠
2. Draw OF to form ∆FOE
3. ∠ 1 is an exterior angle of ∆FOE
4. m∠ 1 = x + y
5. OEOF ≅
6. ∆FOE is an isosceles triangle
7. x = y
8. m∠ 1 = x + x = 2x
9. 2x = m∠ 1, x = ½ m∠ 1
10. But ∠ 1 is a central angle
11. m∠ 1 = m DF
12. x = ( )mDFDEFm
2
1
=∠
1. Given
2. Line determination postulate
3. Definition of exterior angle
4. Exterior angle theorem
5. Radii of the same circle are congruent
6. Definition of isosceles triangle
7. Base angles of isosceles triangle are
congruent
8. Substitution (Steps 4 and 7)
9. Multiplication property of equality
10. Definition of central angle
11. Measure of a central angle equals its
intercepted arc.
So, we have proven case 1. Let us now prove case 2 of the inscribed angle theorem.
Case 2. (The center of the circles lies in the interior of the inscribed angle)
Given : Circle O with inscribed PQR∠
Prove: m PQR∠ =
2
1
m PR
O
●
O
Q
P
S
R
a
b
D
E
F
●
1
x
y
O

16
Proof:
Statements Reasons
1. Circle O with inscribed PQR∠ . Use the
given notation in the figure.
2. Draw diameter QS
3. m PQR∠ = a + b
4. a = mPS2
1
b = mSR2
1
5. a + b = mPS2
1
+ mSR2
1
= )(2
1
mSRmPS +
6. mPR = mPS + mPR
7. m `PQR∠ =
2
1
(mPS + mPR)
8. m `PQR∠ = 2
1
mPR
1. Given
2. Line determination Postulate
3. Angle Addition Postulate
4. Inscribed angle theorem (Case 1)
5. Addition Property of Equality
6. Arc Addition Postulate
7. Transitive Property of Equality
Case 3. (The center is in the exterior of the inscribed angle)
Given: BAC∠ is an inscribed angle in circle O
Use the additional notation in the figure
Prove: mBCBACm 2
1
=∠
Proof:
Statements Reasons
1. Draw diameter AD
2. BACmDABmDACm ∠+∠=∠
3. DABmDACmBACm ∠−∠=∠
4. mDCDACm 2
1
=∠
mDBDABm 2
1
=∠
5. mDBmDCBACm 2
1
2
1
−=∠ = 2
1
(mDC-mDB)
6. mDC = mDB + mBC
7. mBC = mDC – mDB
8. mBCBACm 2
1
=∠
2. Angle Addition Postulate
3. Subtraction Property of Equality
4. Inscribed angle Theorem (Case 1)
5. Substitution
8. Substitution
From the proofs that were given, we can therefore conclude that wherever in the
circle the inscribed angle is located, it is always true that its measure is one-half its
intercepted arc.
A
C
B
D●
O
a
x

17
Examples. Use the figure at the right.
1. Given: circle O. 80=∠BODm
Find: mBD, BADm∠
Solution:
Since 80=∠BODm , then
a. mBD = 80
b. BADm∠ = BD2
1
= )80(2
1
= 40
2. Given: circle O. 37=∠BADm
Find: mBD , BODm∠
Solution:
mBDBADm 2
1
37 ==∠
mBD = 2(37) = 74
mBDBODm =∠
74=∠BODm
Like in the study of central angles and its measure, discussing inscribed angles and
its measure also involves many theorems. Each previous theorem studied is always a tool
in proving the next theorem.
The following theorem is one of the most useful theorem in solving problems which
involve inscribed angles.
Theorem: Angle in a semicircle theorem.
An angle inscribed in a semicircle is a right angle.
Given: Circle O. BAC is a semicircle.
Prove: BAC∠ is a right angle. ( 90=∠BACm
O
A
B
D
O
●B
A
C

18
Proof:
Statements Reasons
1. Draw BC passing through center O.
2. ∠ ABC, ∠ ACB, and ∠ BAC are all inscribed
angles.
3. ACABCm 2
1
=∠ , ABACBm 2
1
=∠
4. mBAC = mAC + mAB
5. BAC is a semicircle
6. mBAC = 180
7. mAC + mAB = 180
8. ABACACBmABCm 2
1
2
1
+=∠+∠ = )(2
1
ABAC +
9. ACBmABCm ∠+∠ = 2
1
(180) = 90
10. ACBmABCm ∠+∠ + BACm∠ = 180
ACBmABCm ∠+∠ = 90
11. BACm∠ = 90
12. BAC∠ is right angle
1. Definition of diameter
2. Definition of inscribed angles
3. Inscribed Angle Theorem
5. Given
6. The measure of a semicircle is 180
8. Addition Property of Equality (Step 3)
10. The sum of the angles of a triangle
is 180.
(Step 10 – step 9)
12. Definition of a right angle
From this point onward, you can use this very important theorem in proving or in
exercises.
There are other theorems on inscribed angle that are also important as the previous
theorem. Of those theorems, we will prove two and the rest, you can answer as exercises.
Theorem:
Inscribed angles subtended by the same arc are congruent.
Given: Circle O. MN subtends both ∠ T and ∠ P
∠ T and ∠ P are inscribed angles
Prove: ∠ T ≅ ∠ P
Proof:
Statements Reasons
1. In circle O, MN subtends both ∠ T and
∠ P. ∠ T and ∠ P are inscribed angles.
2. mMNTm 2
1
=∠
mMNPm 2
1
=∠
3. PmTm ∠=∠
4. PT ∠≅∠
1. Given
4. Definition of congruent angles
● O
M
T
P N

19
The next theorem is about polygon inscribed in a circle.
Definition:
A polygon inscribed in a circle is polygon whose vertices lie on the circle.
Examples: The figures below show examples of inscribed polygon.
Inscribed triangle Inscribed Inscribed Inscribed
Quadrilateral Pentagon Hexagon
Theorem: Opposite angles of a inscribed quadrilateral are supplementary.
Given: Circle A. PRST is an inscribed quadrilateral.
Prove: ∠ P and ∠ S are supplementary
∠ R and ∠ T are supplementary
Proof:
Statements Reasons
1. Circle A. PRST is an inscribed quadrilateral.
2. ∠m P = ½ mRST
∠m S = ½ mRPT
∠m R = ½ mPTS
∠m T = ½ mPRS
3. ∠m P + ∠m S = ½ mRST + ½ mRPT
4. ∠m P + ∠m S = ½( mRST + mRPT)
5. mRST + mRPT = 360
6. ∠m P + ∠m S = ½(360)
7. ∠m P + ∠m S = 180
8. ∠ P and ∠ S are supplementary
9. ∠m R + ∠m S = ½ mPTS + ½ mPRS
10. ∠m R + ∠m S = ½ (mPTS + mPRS)
11. mPTS + mPRS = 360
12. ∠m R + ∠m S = ½(360)
13. ∠m R + ∠m S = 180
14. ∠ R and ∠ T are supplementary
1. Given
2. Inscribed angle theorem
3. Addition property of equality
4. Factoring
5. The arc of the whole circle is 360º
7. Algebraic process (step 6)
8. Definition of supplementary angles
9. Addition property of equality
10. Factoring
11. The arc of the whole circle is 360º
13. Algebraic process (step 6)
14. Definition of supplementary angles
A
●
R
S
P
T
●
● ● ●

20
Examples:
1. Given: XY is a diameter.
a. What kind of angle is ∠ Z?
b. If ∠m X = 35, what is ∠m Y?
c. If ∠m Y = 73, what is mXZ? What is mYZ?
Answers:
a. Since XY is a diameter, then XZY is
a semicircle and ∠ Z is inscribed in a semicircle.
Therefore, ∠ Z is a right angle.
b. m∠ X + m∠ Y = 90.
m∠ Y = 90 - m∠ X
m∠ Y = 90 – 35
m∠ Y = 65
c. ∠ Y intercepts XZ.
m XZ = 2(75) = 150
m YZ = 180 – 150
m YZ = 30
2. MNOP is inscribed in circle E. If m M∠ = 94,
what is m∠ O?
Answer:
∠ M and ∠ O are supplementary.
m∠ M +m∠ O = 180
m∠ O = 180 - m∠ M
= 180 – 94
= 86
3. Given: Circle O. AB is a diameter
∠m 1 = 36 and ∠m 3 = 61.
Find: ∠m 2, ∠m 4, ∠m CBD,
∠m ADB, ∠m ACB, mCBD,
∠m CAD, mAD
Solution:
∠m 1 = 36, mAC = 2(36) = 72
∠m 3 = 61, mBD = 2(61) = 122
∠m 2 = ½ AD
mAD = 180 – BD
= 180 – 122 = 58
∠m 2 = ½ (58)
= 29
M
N
P
O
●
E
C
BA
D
O
● 1
2
4
3
A
●X
Z
Y

21
∠m 4 = ½ CB
mCB = 180 – mAC
= 180 – 72
= 108
∠m 4 = ½ (108)
= 54
∠m CBD = ½(m AC + mAD)
= ½(72 + 58)
= ½(130)
= 65
∠m ADB = 90 (Angle in a semicircle)
∠m ACB = 90 (angle in a semicircle)
mCBD = mCB + mBD
= 108 + 122
= 230
∠m CAD = ½(mCBD)
= ½ (230)
= 115
Try this out
A. Given: AB is a diameter of circle O.
79=mAC .
Find:
1. ∠m AOC
2. ∠m ABC
3. ∠m COB
B. Given: Circle A., XY and BE are diameters
∠m XAE = 104.
Find:
4. m XE
5. m BX
6. ∠m E
7. ∠m B
8. ∠m BXY
9. ∠m YXE
A
C
O
B
B
X
A
E
Y

22
C. Using the given figure, find:
10. x
11. ∠m MNQ
12. ∠m MOQ
13. ∠m POQ
14. ∠m M
15. ∠m MON
D. BD is a diameter of circle A.
If m BC = 78, and m DE = 132,
find:
16. m CD 23 ∠m 6
17. m BE 24. ∠m 7
18. ∠m 1 25. ∠m 8
19. ∠m 2 26. ∠m 9
20. ∠m 3 27. ∠m 10
21. ∠m 4
22. ∠m 5
E. PRST is inscribed in circle A.
If ∠m T = (5x – 4)º and ∠m R = (4x + 13)º
find:
28. x
29. ∠m T
30. ∠m R
F. ∆XYZ is inscribed in the circle.
If XZXY ≅ , prove that ∠m X = b - a
QM
O
3x
N P
● D
C
A
B
E
3 4
1
2
9 10
5
6
3
8
7
P
T
S
RA
●
Y
X
Z
a b
●

23
Let’s Summarize
1. A circle is the set of all points that are at the same distance from a given point in the
plane.
2. Some of the lines associated with circle are the following:
a. Radius
b. Chord
c. Diameter
d. Secant
e. Tangent
3. If a radius is perpendicular to a chord, then it bisects the chord.
4. If a radius of a circle bisects a chord that is not a diameter, then it is perpendicular to
the chord.
5. Congruent circles are circles that have congruent radii. Concentric circles are circles
having the same center.
6. A circle is made up of arcs classified as minor arc, semicircle and major arc.
7. A central angle is an angle on the circle whose vertex is the center of the circle.
8. The measure of the central is numerically equal to its intercepted arc.
9. If two minor arcs of a circle or of congruent circle are congruent, then,
a. the corresponding central angles are congruent,
b. the corresponding subtended chords are congruent
10.An inscribed angle is an angle on the circle whose vertex is a point on the circle.
11.The measure of an inscribed angle is equal to one-half its intercepted arc.
12. An angle inscribed in a semicircle is a right angle.
13. Inscribed angle subtended by the same arc are congruent.
14.The opposite angles of an inscribed quadrilateral are supplementary.

24
What have you learned
Answer as indicated.
1. If the diameter of a circle is 15 cm, what is the length of the radius?
2. A line that intersects a circle at one point is called _________.
3. If a radius bisects a chord which is not a diameter, then its is _________ to the
chord.
4. CD is a diameter of circle A. CED is a ______________.
5. CE is a _____________.
6. CDE is a ______________
7. PTON ⊥ at E. If OE = 6 cm, and the radius
of the circle is 10 cm, what is the length of PT ?.
8. AC is a diameter of circle O. Using the given in the
given figure, find
a. ∠m A
b. ∠m C
c. m AB
d. m BC
9. PT is a diameter of circle Q. Find
a. ∠m PQR
b. ∠m RQT
10. ∆ABC is inscribed in circle O.
If the ratio of ∠m A: ∠m B: ∠m C =
2:3:5, Find
a. m BC
b. m AC
c. m AB
A
D
C
E
●
O
TP
N
E
A
BC
●A
RQ
P
T
2x
3x
A
C
B
● O

25
Answer Key
1. a. circle O
b. MN
c. MO , ON
d. MT , MR
e. MR
f. MS
2. bisects
3. minor arc
4. major arc
5. 4 cm
6. 5 cm
7. 93°
8. 46.5°
9. 98°, 49°
10. 77°
Try this out
Lesson 1
A. 1. circle O
2. AC , BD
3. ,AD BC
4. EC , BC
5. CF
B. 1. 12 cm 5. 119 cm
2. 2 6 cm 6. 35 cm
3. 171 cm 7. 5 cm
4. 43 cm 8. 29 cm
C. 1. 149 cm 5. 55 cm
2. 210 cm 6. 11 cm
3. 269 cm 7. 15 cm
4. 2 41 cm 8. 14 cm

26
D. 1. 114 cm 5. 76 cm
2. 312 cm 6. 22 cm
3. 58 cm 7. 20 cm
4. 192 cm 8. 16
Problem Solving:
1. ME = 52, PE = 52, MP = 104
2. maximum distance is 12 cm
minimum distance is 2 cm
Lesson 2
A. 1. 180º 3. 98º
2. 82º 4. 278º
5. 262º
B. 1. 73º 5. 73º
2. 73º 6. 107º
3. 107º 7. 253º
4. 107º
C. 1. 29º 5. 29º
2. 87º 6. 87º
3. 29º 7. 116º
4. 87º
D. 1. 56
2. 56
3. AB and CD
E. 1. Each arc measures 120
2. AC = 86º
3. ABC = 274º
F. PQ = 90
QR = 120
PR = 150
G. 1. 38º 5. 38º
2. 157º 6. 157º
3. 76º 7. 76º
4. 89º 8. 89º

27
H. 1. 35º 5. 110º
2. 35º 6. 110º
3. 35º 7. 17º
4. 110º
Lesson 3
A. 1. 79 3. 101
2. 39.5
B. 4. 104 7. 52
5. 76 8. 52
6. 38 9. 38
C. 10. 36 13. 108
11. 36 14. 36
12. 72 15. 108
D. 16. 102 22. 24
17. 48 23. 39
18. 51 24. 66
19. 66 25. 24
20. 39 26. 105
21. 51 27. 75
E. 28. 19
29. 91
30. 89
F. Proof:
1. ∠ X + ∠ Y + a = 180 1. Sum of the measures of the angles of a
triangle is 180.
2. b = m∠ X + m∠ Y 2. Exterior angle theorem
3. m∠ Y = a 3. Angles opposite equal sides in the same
triangle are congruent
4. b = m∠ X + a 4. Substitution
5. m∠ X = b – a 5. Subtraction Property of Equality
1. 7.5 cm 8. a. 22.5 10. a. 72
2. tangent b. 67.5 b. 108
3. perpendicular c. 135 c. 180
4. semicircle d. 45
5. minor arc 9. a. 72
6. major arc b. 108
7. 16 cm

Module 2
Circles
This module will discuss in detail the characteristics of tangent and secants; the
relationship between tangent and radius of the circle; and how secant and tangent in a
circle create other properties particularly on angles that they form. This module will also
show how the measures of the angles formed by tangents and secants can be determined
and other aspects on how to compute for the measures of the angles.
This module is written for you to
1. Define and illustrate tangents and secants.
2. Show the relationship between a tangent and a radius of a circle.
3. Identify the angles formed by tangents and secants.
4. Determine the measures of angles formed by tangents and secants.
If CB and CD are tangents to circle A, then
1. CB ___ CD
2. ABCB ____
3. CB and CA are tangents to circle O.
If 160=∠BOAm , then =∠Cm _____.
4. If 22=∠BCOm , what is ACOm∠ ?
A
B
C
D
C
B
O
A

2
5. In the figure, if m PTA = 242, what is PALm∠ ?
6. Two secants GD and BL intersect at A.
If m BG = 83 and m LD = 39, find GABm∠ .
7. In the figure, if m MX = 54, and mAX = 120, what is Nm∠ ?
8. AC and AT are tangents to the circle
with C and T as the points of tangency.
If ACT∆ is an equilateral triangle,
find m CT.
9. AC and AT are secants. If 23=∠Am
and m CT = 66, find m BM.
A
T
BL
P
O
G
L
D
A
B
N
X
M
A
A
C
T
M
A
B
M
C
T

3
10. DB is a diameter of circle O. If the ratio
of DE:EB is 3:2, what is Xm∠ ?
What you will do
Lesson 1
Circles, Tangents, Secants and Angles They Form
A line on the same plane with a circle may or a)
may not intersect a circle. If ever a line intersects a
circle, it could be at one point or at two points.
The figures at the right showed these three instances.
Figure a showed a line that does not
intersect the circle.
b)
Figure b showed that line t intersects the
circle at only one point.
Figure c showed line l intersecting the
circle at two points A and C.
c)
We will focus our study on figures b and c.
In figure b, line t is called a tangent and point B
is called the point of tangency. Therefore a tangent is a
line that intersects a circle at only one point and the point
of intersection is called the point of tangency.
In figure c, line l intersects the circle at two points A and C. Hence, line l is called
a secant. Thus a secant is a line that intersects a circle at two points.
Some properties exist between tangent and circle and they will be discussed here in
detail. The first theorem is given below.
Theorem: Radius-Tangent Theorem. If a line is tangent to a circle, then it is perpendicular
to the radius at the point of tangency.
t B
A
C
l
X
E
D
O
B

4
Given: line t is tangent to circle O at A.
OA is a radius of the circle.
Prove: t ⊥ OA
Proof:
Statements Reasons
1. Let B be another point on line t.
2. B is on the exterior of circle O
3. OA < OB
4. ⊥OA t
1. The Line Postulate
2. Definition of a tangent line ( A tangent
can intersect a circle at only one point .
3. The radius is the shortest segment
from the center to the circle and B is on
the exterior of the circle.
4. The shortest distance from a point to a
line is the perpendicular segment.
Example:
In the figure, if AC is tangent to
circle B, then
BDAC ⊥ at D.
The converse of the theorem is also true.
Converse: The line drawn perpendicular to the radius of a circle at its end on the circle is
tangent to the circle.
Illustration:
If BDAC ⊥ at D, then
AC is tangent to circle B.
Examples:
GY is tangent to circle A.
1. What kind of triangle is ∆ AGY? Give reason.
2. If 79=∠Am , what is Ym∠ ?
Solutions:
1. ∆AGY is a right triangle because GY is tangent to circle A and tangent line is
perpendicular to the radius of the circle. Perpendicular lines make right angles between
them thus AGY∠ is a right angle making ∆AGY a right triangle.
O
A B
t
B
DC A
B
DC A
A
Y
G

5
2. Since ∆AGY is a right triangle, then
∠m A + ∠m Y = 90
79 + ∠m Y = 90
∠m Y = 90 – 79
= 11
A circle is composed of infinite number of points, thus it can also have an infinite
number of tangents. Tangents of the same circle can intersect each other only outside the
circle.
At this point, we will discuss the relationship of tangents that intersect the same
circle. As such, those tangents may or may not intersect each other. Our focus here are
those tangents that intersect each other outside the circle.
Consider the given figure:
AM and AY are tangent segments
from a common external point A. What
relationship exists between AM and AY ?
The next theorem will tell us about this
relationship and other properties related to
tangent segments from a common external point.
Theorem: If two tangent segments are drawn to a circle from an external point then
a. the two tangent segments are congruent and
b. the angle between the segments and the line joining the external point and the
center of the circle are congruent.
Given: Circle A. BC and BD are two tangent segments
from a common external point B. C and D are
the points of tangency.
Prove: a. BDBC ≅
b. DBACBA ∠≅∠
Proof:
Statements Reasons
1. Draw AC , AD , AB
2. BC and BD are two tangent
segments from a common external
point B.
3. BCAC ⊥ , BDAD ⊥
4. ACB∠ and ADB∠ are right angles
2. Given
3. A line tangent to a circle is perpendicular
to the radius at the point of tangency
4. Definition of right angles
O •
A
M
Y
A
C
D
B

6
5. ∆ACB and ∆ADB are right triangles
6. ADAC ≅
7. BDBC ≅
8. ∆ACB ≅ ∆ADB
9. BDBC ≅
10. DBACBA ∠≅∠
5. Definition of right triangles
6. Radii of the same circle are congruent
7. Reflexive property of Congruency
8. Hy L Congruency Postulate
9. Corresponding parts of congruent triangles
10. are congruent.
Examples:
a) In the figure, CB and CD are tangents
to circle A at B and D.
1. If CB = 10 what is CD?
2. If 49=∠BACm , what is BCAm∠ ?
3. ,73=∠BCDm what is DCAmBCAm ∠∠ ?
Solution:
1. Since CB and CD are tangents to the same circle from the same external point,
then CB ≅ CD , and therefore, CB = CD. Thus if CB = 10 then CD = 10
2. 90=∠+∠ BCAmBACm
49 + 90=∠BCAm
4990 −=∠BCAm
= 41
3. )(2
1
BCDmBCAm ∠=∠
= )73(2
1
= 36.5
DCABCA ∠≅∠
5.36=∠=∠ DCAmBCAm
b) PQ, QR and PR are tangents to circle A
at S, M and T respectively. If PS = 7,
QM = 9 and RT = 5, what is the perimeter
of ∆PQR?
Solution:
Using the figure and the given information, It is
therefore clear that PS = PT, QS = QM and
RM = RT.
PQ = PS + SQ
QR = QM + MR
PR = PT + RT
C
B
D
A
P
Q
R
S
T
M
A

7
Perimeter of ∆PQR = PQ + QR + PR
= (PS + SQ) + (QM + MR) + (PT + RT)
= (PS + QM) + (QM + RT) + (PS + RT)
= 2PS + 2QM + 2RT
= 2(PS + QM + RT)
= 2( 7 + 9 + 5)
= 2 (21)
= 42
Every time tangents and secants of circles are being studies, they always come with
the study of angles formed between them. Coupled with recognizing the angles formed is
the knowledge of how to get their measures. The next section will be devoted to studying
angles formed by secants and tangents and how we can get their measures.
Angles formed by secants and tangents are classified into five categories. Each
category is provided with illustration.
1. Angle formed by secant and tangent intersecting
on the circle. In the figure, two angles of this type
are formed, FED∠ and FEB∠ . Each of these angles
intercepts an arc. FED∠ intercepts EF and FEB∠
intercepts EGF.
2. Angle formed by two tangents. In the figure, E∠ is
formed by two tangents. The angle intercepts the
whole circle divided into 2 arcs, minor arc FD , and
major arc FGD.
3. Angle formed by a secant and a tangent that
intersect at the exterior of the circle. C∠ is an
angle formed by a secant and a tangent that
intersect outside the circle. C∠ intercepts two
arcs, DB and AD.
4. Angle formed by two secants that intersect in the
interior of the circle. The figure shows four angles
formed. ,MAN∠ ,NAR∠ PAR∠ and PAM∠ . Each
of these angle intercepts an arc. ,MAN∠ intercepts
MN, ,NAR∠ intercepts NR, PAR∠ intercepts PR
and MAP∠ intercepts MP.
E
F
B D
G
F
E
D
G
C
D
A
B
M
N
A
RP

8
5. Angle formed by two secants intersecting
outside the circle. E∠ is an angle formed
by two secants intersecting outside the
circle. E∠ intercepts two arcs namely
QR and PR
How do we get the measures of angles illustrated in the previous page? To
understand the answers to this question, we will work on each theorem proving how to get
the measures of each type of angle. It is therefore understood that the previous theorem
can be used in the proof of the preceding theorem.
Theorem: The measure of an angle formed by a secant and a tangent that intersect on the
circle is one-half its intercepted arc.
Given: Circle O. Secant m and tangent t intersect
at E on circle O.
Prove: CECEBm 2
1
=∠
Proof:
Statements Reasons
1. Draw diameter ED. Join DC.
2. tDE ⊥
3. DCE∠ is a right angle
4. DEB∠ is a right angle
5. ∆DCE is a right triangle
6. 1∠m + 902 =∠m
7. =∠+∠ BECmm 1 DEBm∠
8. =∠+∠ BECmm 1 90
9. 21 ∠+∠ mm = BECmm ∠+∠1
10. 1∠m = 1∠m
11. 2∠m = BECm∠
12. 2∠m = 2
1
mCE
13. BECm∠ = 2
1
mCE
2. Radius-tangent theorem
3. Angle inscribed in a semicircle is a right
angle.
4. Perpendicular lines form right angles
5. Definition of right triangle
6. Acute angles of a right triangle are
complementary
7. Angle addition Postulate
8. Definition of complementary angles
10. Reflexive Property of Equality
12. Inscribed angle Theorem
13. Substitution
Illustration:
In the given figure, if mCE = 104, what is the m∠ BEC? What is m CEF∠ ?
Q
P
S
R
E
BE
C
D
F
O
t

9
Solution:
BECm∠ = 2
1
mCE
BECm∠ = 2
1
(104)
= 52
m CEF∠ = ½ (mCDE)
m CEF∠ = ½ (360 – 104)
= ½ (256)
= 128
Let’s go to the next theorem.
Theorem: The measure of an angle formed by two tangents from a common external point
is equal to one-half the difference of the major arc minus the minor arc.
Given: Circle O. AB and AC are tangents
Prove: )(2
1
BCBXCAm −=∠
Proof:
Statements Reasons
1. Draw chord BC
2. In ∆ABC, 1∠ is an exterior angle
3. 1∠m = 2∠m + Am∠
4. Am∠ = 1∠m - 2∠m
5. 1∠m = ½ mBXC
2∠m = ½ mBC
6. Am∠ = ½ mBXC - ½ mBC
7. Am∠ = ½( mBXC – m BC)
3. Exterior angle theorem
5. Measure of angle formed by secant and
tangent intersecting on the circle is one-half
the intercepted arc.
6. Substitution
7. Algebraic solution (Common monomial
Factor)
Illustration:
Find the Am∠ if mBC = 162.
Solution:
Since Am∠ = ½( mBXC –m BC) then we have to find first the measure of
major arc BXC. To find it, use the whole circle which is 360o
.
A
B
C
1
2
X
O

10
mBXC = 360 – mBC
= 360 – 162
= 198
Then we use the theorem to find the measure of A∠
Am∠ = ½( mBXC –m BC)
= ½ (198 – 162)
= ½ (36)
Am∠ = 18
We are now into the third type of angle. Angle formed by secant and tangent
intersecting on the exterior of the circle.
Theorem: The measure of an angle formed by a secant and tangent intersecting on the
exterior of the circle is equal to one-half the difference of their intercepted arcs.
Given: BA is a tangent of circle O
BD is a secant of circle O
BA and BD intersect at B
Prove: )(2
1
ACADBm −=∠
Proof:
Statements Reasons
1. BA is a tangent of circle O, BD is a
secant of circle O
2. Draw AD
3. 1∠ is an exterior angle of ∆ DAB
4. ADBmBmm ∠+∠=∠1
5. ADBmmBm ∠−∠=∠ 1
6. =∠1m ½m AD
7. ADBm∠ = ½ mAC
8. Bm∠ = ½ mAD – ½ mAC
9. Bm∠ = ½ (mAD– mAC)
1. Given
4. Exterior angle Theorem
6. The measure of an angle formed by
secant and tangent intersecting on the circle
equals one-half its intercepted arc.
7. Inscribed angle Theorem
8. Substitution
9. Simplifying expression
Illustration:
In the figure if mAD = 150, and mAC = 73, what is the measure of B∠ ?
A
B
C
O
D
1

11
Solution:
Bm∠ = ½ (mAD– mAC)
= ½ (150 – 73)
= ½ (77)
Bm∠ = 38.5
The next theorem will tell us how angles whose vertex is in the interior of a circle can
be derived. Furthermore, this will employ the previous knowledge of vertical angles whether
on a circle or just on a plane.
Theorem: The measure of an angle formed by secants intersecting inside the circle equals
one-half the sum of the measures of the arc intercepted by the angle and its vertical angle
pair.
Given: AC and BD are secants intersecting inside
circle O forming 1∠ with vertical angle pair
.CED∠ (We will just work on one pair of
vertical angles.)
Prove: )(1 AEBmm ∠∠ = ½ (AB + DC)
Proof:
Statements Reasons
1. AC and BD are secants intersecting
inside circle O.
2. Draw AD
3. 1∠ is an exterior angle of ∆AED
4. 1∠m = DACm∠ + ADEm∠
5. DACm∠ = ½ mDC
ADEm∠ = ½ mAB
6. 1∠m = ½ mDC + ½ mAB
1∠m = ½ (mDC + mAB)
1. Given
6. Substitution
Illustration:
Using the figure, find the measure of 1∠ if mAB = 73 and mCD = 90.
Solution:
Using the formula in the theorem,
1∠m = ½ (mDC + mAB)
= ½ ( 90 + 73)
= ½ (163)
= 81.5
E
A
B
C
O
D
1
E

12
Let us discuss how to find the measure of the angle formed by two secants
intersecting outside the circle.
Theorem: The measure of the angle formed by two secants intersecting outside the circle is
equal to one-half the difference of the two intercepted arcs.
Given: AB and CD are two secants
intersecting outside of circle O
forming BEC∠ outside the circle.
Prove: BECm∠ = ½ (AD – BC)
Proof:
Statements Reasons
1. AB and CD are secants of circle O
forming BEC∠ outside the circle.
2. Draw DB
3. 1∠ is an exterior angle of ∆ DBE
4. 1∠m = 2∠m + BECm∠
5. BECm∠ = 1∠m - 2∠m
6. 1∠m = ½ mAD
2∠m = ½ mBC
7. BECm∠ = ½ mAD – ½ mBC
BECm∠ = ½ (mAD – mBC)
1. Given
3. Definition of exterior angle of a triangle
7. Substitution
Illustration:
Find the measure of ∠ E if mAD = 150 and mBC = 80.
Solution:
Again we apply the theorem using the formula:
BECm∠ = ½ (mAD – mBC)
= ½ (150 – 80
= ½ ( 70)
= 35
B
E
D
C
A
1
2

13
Example 1.
In each of the given figure, find the measure of the unknown angle (x).
1. 2.
3. 4.
5.
Solutions:
1. Given: AB = 1500
Find: xm∠
x∠ is an angle formed by a secant and a tangent whose vertex is on the circle. x∠
intercepts AB.
xm∠ = ½ AB
xm∠ = ½ (150)
xm∠ = 75
A
B
x
1500
M
PN
O
1570
x
O
x
A
M
Y
P
●
780
x
F
D
E
G
H
670
400
R
P
Q
T
S
1060
380

14
2. Given: m MP = 157
Find: xm∠
x∠ is an angle formed by two tangents from a common external point. x∠ intercepts
minor arc MP and major arc MNP
xm∠ = ½ ( MNP – MP)
m MNP + mMP = 360
mMNP = 360 – m MP
= 360 – 157
mMNP = 203
xm∠ = ½ (203 – 157)
= ½ (46)
= 23
3. Given: m AP = 78
AY is a diameter
Find: xm∠
Since AY is a diameter, then AY is a semicircle and m AY = 180. Therefore
a. m AP + m PY = 180
m PY = 180 – m AP
m PY = 180 – 78
m PY = 102
b. xm∠ = ½ ( mPY – mAP)
= ½ ( 102 – 78)
= ½ (24)
= 12
4. Given: mFD = 67, m GE = 40
Find: xm∠
x∠ is an angle formed by secants that intersect inside the circle, Hence
xm∠ = ½ (mFD + mGE)
= ½ (67 + 40)
= ½ (107)
= 53.5

15
5. Given: mSR = 38, mPQ = 106
Find: xm∠
x∠ is an angle formed by two secants whose vertex is outside the circle. Thus
xm∠ = ½ (mPQ – mSR)
= ½ (106 – 38)
= ½ (68)
= 34
Example 2:
Find the unknown marked angles or arcs (x and y) in each figure:
1. 2.
3. 4.
5. 6.
M D
B CA
xy
●
QP
R
N
2450
y
y
x
R S
U
T
580320 x220y
P
Q
S
R
300
A
B C
D
E
x
350
E
F
D
G
H
270
370
x

16
Solutions:
1. Given : mBMD = 210
Find: xm∠ , ym∠
mBMD + mBD = 360 (since the two arcs make the whole circle)
mBD = 360 – mBMD
= 360 – 210
= 150
a. xm∠ = ½ mBD
= ½ (150)
= 75
b. ym∠ = ½ m BMD
= ½ (210)
= 105
2. Given: mPNR = 245
Find: xm∠ , ym∠
mPNR + mPR = 360 (since the two arcs make a whole circle)
mPR = 360 – mPNR
= 360 – 245
= 115
But xm∠ = mPR (Central angle equals numerically its intercepted arc)
xm∠ = 115
ym∠ = ½ (mPNR – mPR)
= ½ (245 – 115)
= ½ (130)
= 65
3. Given: mRU = 32, mST = 58
Find: xm∠ , ym∠
xm∠ = ½ (mRU + mST)
= ½ (32 + 58)
= ½ (90)
= 45

17
xm∠ + ym∠ = 180 Linear Pair Postulate
ym∠ = 180 - xm∠
ym∠ = 180 – 45
= 135
4. Given: mQR = 30, 1∠m = 22 To check;
Find : xm∠ , y 1∠m = ½ (mPS + mQR)
First, find y by using m 1∠ = ½ (14 + 30 )
1∠m = ½ ( y + 30) = ½ ( 44)
22 = ½ (y + 30) = 22
2(22) = y + 30
44 = y + 30
y = 44 – 30
y = 14
5. Given: Am∠ = 35, mCD = 110
Find: x To check:
Am∠ = ½ (110 – x) Am∠ = ½ (110 – x)
35 =
2
110 x−
Am∠ = ½ (110 – 40)
2(35) = 110 – x Am∠ = ½ (70)
70 = 110 – x Am∠ = 35
x = 110 – 70
x = 40
6. Given: Em∠ = 27, mDH = 37
Find: x To check:
Em∠ = ½ (x – 37) Em∠ = ½ (x – 37)
27 =
2
37−x
Em∠ = ½ (91 – 37)
54 = x – 37 Em∠ = ½(54)
x = 54 + 37 Em∠ = 27
x = 91

18
Try this out.
1. BC is tangent to circle O at A. mDE = 68
mAF = 91. Find
a. mEA
b. mDF
c. m∠ BAE
d. m∠ EAD
e. m DAF∠
f. FACm∠
2. ∆DEF is isosceles with DFDE ≅ .
∠m 1 = 82. Find
a. ∠m D
b. mDE
c. mEF
d. ∠m 2
e. ∠m 3
3. If x = 18 and y = 23,
find 1∠m .
4. If mDE = 108 and m DOC∠ = 85, find:
a. mEA
b. ∠m EAF
c. ∠m DAC
d. ∠m CAB
e. ∠m 1
5. Using the given figure, find x and y.
O
F
H
E
G
2
3
D
1
B
C
A
1
y
x
D
D
E
O C
AF B
1
x
1100
600
y
D A
E
F
●
B
C
680
910
O

19
6. EC is tangent to circle O. AB is a diameter.
If mDB = 47, find mAD, m ECD∠
7. A polygon is said to be circumscribed about a
circle if its sides are tangent to the circle.
∆PRT is circumscribed about circle O.
If PT = 10, PR = 13 and RT = 9, find
AP, TC and RB.
8. PT is tangent to circle O at P.
If mNP = 90, and mMXP = 186, find
a. ∠m 1
b. ∠m 2
c. ∠m 3
d. ∠m 4
e. ∠m 5
f. ∠m 6
9. O is the center of the given circle.
If mBD = 122 find
a. ∠m 1 d. ∠m 4
b. ∠m 2 e. ∠m 5
c. ∠m 3 f. ∠m 6
A
D
C
B
O
E
P
T
A
B
C
R
O
4
3
1
M
TP
N
X 2
5
6
3
E
D
F
O
A B C
5
1
4
6
2

20
Let’s Summarize
1. A tangent is a line that intersect a circle at only one point.
2. A. secant is a line that intersect a circle at two points.
3. If a line is tangent to a circle, it is perpendicular to a radius at the point of tangency.
4. If two tangents are drawn from an exterior point to a circle then
a) the two tangent segments are congruent
b) the angle between the segment and the line joining the external point and the
center of the circle are congruent.
5. The measure of an angle formed by a secant and a tangent intersecting on the circle
is equal to one half the measure of the intercepted arc.
6. The measure of an angle formed by two tangents from a common external point is
equal to one-half the difference of the measures of the intercepted arcs.
7. The measure of an angle formed by secant and tangent intersecting outside the
circle is equal to one-half the difference of the measures of the intercepted arcs.
8. The measure of an angle formed by two secants intersecting inside the circle is equal
to one-half the sum of the measure of the intercepted arc of the angle and its
vertical angle pair.
9. The measure of an angle formed by two secants intersecting outside the circle is
equal to one-half the difference of the measures of the intercepted arcs.

21
1. QP is tangent to circle O at P. If POQm∠ = 73,
what is Qm∠ ?
2. DE , EF and DF are tangents to circle M.
If DB =- 5, EC = 7 and AF = 4, what is the
perimeter of ∆DEF?
3. PS AND PQ are secant and tangent of
circle A. If mRQ = 52, what is Pm∠ ?
4. Given circle E with secants AB and CD .
If mBD = 53 and mBC = 117, find
BEDm∠ .
5. XY is tangent to circle at A . If mAB = 105, and
AB ≅ BP, find BAPm∠ and
6. PAYm∠
O
Q
P
E
D
F
B
A
C
A
S
Q
P
R
EA
D
B
C
P
B
AX Y

22
7. Given circle A with secants OQ and OP .
If mRS = 32 and mQR = 2mRS, find
Om∠ ?
8. AB ║ DE . BE is tangent to the circle
at B and intersects DE at E.
If mAB = 110 and mAD = 70, then
ABFm∠ = ___________.
9. BECm∠ = __________.
10. Using the same figure, if Em∠ = 42, mBC = 60, find mAB.
A
O
SP
R
Q
A
F
B
E
CD

23
Answer Key
1. ≅ or congruent
2. ⊥ or perpendicular
3. 40
4. 22
5. 59
6. 61
7. 33
8. 120
9. 20
10. 18
Try this out
Lesson 1
1. a. 112
b. 89
c. 56
d. 34
e. 44.5
f. 45.5
2. a. 16
b. 164
c. 32
d. 82
e. 16
3. 41
4. a. 72
b. 36
c. 42.5
d. 47.5
e. 54

24
5. x = 70
y = 50
6. mAD = 133
m ECD∠ = 43
7. AP = 7
TC = 3
RB = 6
8. a. 45
b. 45
c. 42
d. 93
e. 57
f. 48
9. a. 122
b. 61
c. 29
d. 29
e. 29
f. 45
1. 17
2. 32
3. 38
4. 58
5. 52.5
6. 75
7. 26
8. 55
9. 55
10.84

Module 1
Properties of Quadrilaterals
This module is about the properties of trapezoids and parallelograms. In
this module, you will learn to compute problems involving the median of a
trapezoid, the base angles and diagonals of an isosceles trapezoid, as well as
problems involving the diagonals, angles and sides of parallelograms
This module is designed for you to
1. apply inductive/deductive skills to derive certain properties of a
trapezoid.
• median of a trapezoid
• base angles of an isosceles trapezoid
• diagonals of an isosceles trapezoid
2. apply inductive and deductive skills to derive the properties of a
parallelogram
• each diagonal divides a parallelogram into two congruent
triangles
• opposite angles are congruent
• non-opposite angles are supplementary
• opposite sides are congruent
• diagonals bisect each other.
True or False
1. The median of a trapezoid is twice the sum of the lengths of its bases.
2. The base angles of an isosceles trapezoid are congruent.
3. The diagonals of any trapezoid are congruent.
4. Non-opposite angles of a parallelogram are complementary.

2
ABCD is a trapezoid with median EF.
D 10 C
13
16
A B
5. If DC = 12 cm and AB = 23 cm, what is EF?
6. If EF = 2x + 3, DC = x + 5 and AB = 2(x + 3), what is AB?
Quadrilateral BEST is a parallelogram.
2x + 3
T S
3(x – 1)
B E
7. If BE = 3(x – 1) and TS = 2x + 3, what is BE + TS?
Quadrilateral ABCD is a rhombus.
D C
4x + 10
5x
A B
8. If m ∠A = 5x and m∠C = 4x + 10, what is m∠ADC in degrees?
Quadrilateral ABCD is a parallelogram. Diagonals DB and AC intersect at E.
D C
E
A B

3
9. Find AC if AE = x + 2 and CE = 3x - 6
Quadrilateral TEAM is a parallelogram.
M A
x 3x + 20
T E
10.If m∠T = x and m∠E = 3x + 20 . find m∠A?
What you will do
Lesson 1
The Median of a Trapezoid
A trapezoid is a quadrilateral with exactly one pair of parallel sides. The
median of a trapezoid is the segment joining the midpoints of the non-parallel
sides. In the trapezoid ABCD below, EF is the median. It joins the midpoints E
and F of side AD and BC respectively.
D C
E F
A B
Illustration
If the length of the upper base of a trapezoid is 4 cm, and the lower base is
6 cm, what do you think is the length of the median?
Do the following:
1. Using a ruler draw a segment 4 cm long. Name the segment HT.
2. Draw another segment 6 cm long parallel to segment HT. Name the
segment MA.

4
3. Connect points H and M.
4. Connect points T and A.
5. Using a ruler, carefully determine the midpoints of HM and TA. Name the
midpoints of HM and TA, G and E respectively.
6. Connect points G and E. Carefully measure the length of GE.
What did you discover? Did you discover the following?
1. GE is parallel to HT and MA.
2. GE =
2
1
(HT + MA)
Example 1
ABCD is a trapezoid with median EF. If DC = 8 cm and AB = 14 cm, find EF.
D 8 C
E F
14
A B
Solution:
Step 1. Write the formula.
EF =
2
1
(AB + DC)
Step 2. Substitute the values of DC and AB into the formula. Solve for EF.
EF =
2
1
(14 + 8)
EF =
2
1
(22)
EF = 11
Therefore EF is 11 cm.

5
Example 2
EFGH is a trapezoid with median IJ. If HG = 12 cm and IJ =15 cm, what is
EF?
H 12 cm G
I 15 cm J
E F
IJ =
2
1
(EF + HG)
Step 2. Substitute the values of HG and IJ into the formula. Solve for EF.
15 =
2
1
(EF + 12)
15 = EF + 12
2
EF + 12 = 15(2)
EF = 30 – 12
EF = 18
Therefore EF is 18 cm.
Example 3
ABCD is a trapezoid with median EF. If DC = x + 5, EF = 2x + 1 and
AB =4x – 10, find EF.
D x + 5 C
E 2x + 1 F
A 4x – 10 B

6
Solution:
EF =
2
1
(AB + DC)
Step 2. Substitute the values of EF, AB and DC into the formula. Solve for x.
2x + 1 =
2
1
(4x – 10 + x + 5)
2x + 1 =
2
1
( 5x –5)
2(2x +1) = 5x – 5
4x + 2 = 5x – 5
4x – 5x = -5 – 2
- x = -7
x = 7
Step 3. Substitute 7 for x. Solve for EF
EF = 2x + 1
= 2(7) + 1
= 14 + 1
= 15
Therefore EF = 15 cm.
Try this out
Set A
ABCD is a trapezoid with median EF.
D C
E F
A B
1. If DC = 4 and AB = 10, what is EF?
2. If DC = 7 and AB = 12, what is EF?
3. If AB = 14.5 and DC = 8.5, what is EF?
4. If AB = 10.5 and DC = 6.5, what is EF?
5. IF DC = 10 and EF = 12, what is AB?
6. If DC = 14 and EF = 16, what is AB?
7. If AB = 14.6 and EF = 10.4, what is DC?

7
8. IF AB = 22.8 and EF = 16.2, what is DC?
9. If DE = 14 what is AE?
10.If BF = 24, what is CF?
Set B
EFGH is a trapezoid with median IJ.
H G
I J
E F
1. If IJ = x, HG = 8 and EF = 12, what is x?
2. If IJ = y, HG = 14, and EF = 20, what is Y?
3. If IJ = x + 2, HG = 10 and EF = 14, what is x?
4. If IJ = y + 3, HG = 14 and EF = 18, what is y?
5. If HG = x , IJ = 16, and EF = 22, what is x?
6. If HG = Y, IJ = 24 and EF = 30, what is y?
7. If HG = x – 1, IJ = 21 and EF = 34, what is x?
8. If HG = y – 2, IJ = 20 and EF = 31, what is y?
9. If HI = x + 4 and IE = 6, what is x?
10.If GJ = 10 and FJ = x –4, what is x?
Set C
KLMN is a trapezoid with median OP.
N M
O P
K L
1. If OP = 20, NM = x + 3 and KL = x + 6, what is x?
2. If NM = x – 2, KL = x + 4 and OP = 24, what is x?
3. If OP = 22, NM = x + 4, and KL = x + 8, what is NM?
4. If OP = 24, NM = x – 3 and KL = x + 7, what is KL?
5. If NM = 12, OP = x +3, and KL = x + 10, what is x?
6. If NM = 18, OP = x – 2 and KL = x + 3, what is x?
7. If KL = 30, OP = x + 1 and NM = x – 6, what is OP?

8
D C
B
8. If KL = 34, OP = x - 1, and NM = x – 7, what is NM?
9. If NM = 2x, OP = 3x, and KL = 2(x+5), what is OP?
10.If NM = 2x + 2. OP = 3x + 3 and KL = 2(x + 6), what is NM?
Lesson 2
Isosceles Trapezoid
An isosceles trapezoid is a trapezoid with congruent non-parallel sides.
Trapezoid DAVE below is an isosceles trapezoid. The nonparallel sides ED and
VA are congruent.
E V
D A
Do the following:
1. Using a ruler, draw isosceles trapezoid ABCD with base angles, ∠A and
∠B, on a graphing paper.
A
2. Using a protractor, find the measures of ∠A and ∠B. What do you notice?
3. Using the same protractor find the measures of ∠ D and ∠ C. What do
you notice?
4. Draw the diagonals AC and BD. Using a ruler, find their lengths. Are the
lengths equal?

9
Perhaps you discovered the following properties of an isosceles trapezoid.
2. The diagonals of an isosceles trapezoid are congruent.
If you know the properties of an isosceles trapezoid, you will find
the next set of exercises easy to solve.
Example 1
In isosceles trapezoid ABCD, with base angles, ∠ A and ∠ B. If m∠B = 40,
what is m∠A?
D C
A B
Solution:
Step 1. Base angles of an isosceles trapezoid are congruent.
∠ A ≅∠ B
m∠ A = m∠ B (The measures of congruent angles are equal)
Step 2. Substitute 40 for m∠B
m∠A = 40
Example 2
In isosceles trapezoid ABCD with ∠ A and ∠ B as base angles. If m∠A =
x + 20 and m∠B = 2x. Find m∠A.
D C
A B
Solution:
Step 1. Base angles of an isosceles trapezoid are congruent.
∠ A ≅ ∠ B
m∠ A = m∠ B (the measures two angles congruent are equal)

10
Step 2. Substitute x + 20 for m∠ A and 2x for m∠ B.
x+ 20 = 2x
x – 2x = -20
x = 20
Step 3. Substitute 20 for x in m∠ A = x + 20. Then solve for m∠ A.
m∠ A = x + 20
= 20 + 20
= 40
Example 3
In isosceles trapezoid ABCD, AD = 10. What is BC?
D C
A B
Solution:
Step 1. Non-parallel sides of an isosceles trapezoid are congruent.
BC ≅ AD
BC = AD (Congruent segments have equal lengths)
Step 2. Replace BC with 10
BC = 10
Example 4.
In isosceles trapezoid ABCD, AC = 4x + 4 and BD = 2x + 10. What is x?
D C
A B

11
Solution:
Step 1 The diagonals of an isosceles trapezoid are congruent.
AC ≅ BD
AC = BD ( Congruent segments have equal lengths)
Step 2. Substitute 4x + 4 for AC and 2x + 10 for BD.
4x + 4 = 2x + 10
4x – 2x = 10 – 4
2x = 6
x = 3
Try this out
Set A
PQRS is an isosceles trapezoid
S R
P Q
1. Name the two pairs of base angles.
2. Name the nonparallel sides.
3. Name the parallel sides.
4. If m∠A = 30, what is the m∠B?
5. If m∠B = 60, what is m∠A?
6. If m∠C = 110, what is m∠D?
7. If m ∠D = 105, what is m∠C?
8. If m∠A = 35, what is m∠D?
9. If m∠C = 120, what is m∠B?
10.If m∠A = 35 what is m∠C?
Set B.
EFGH is an isosceles trapezoid.
H G
E F

12
1. If m∠E = x + 10 and m∠F = 50, what is x?
2. If m∠F = x – 15 and m∠E = 70 what is x?
3. If m∠H = x + 20 and m∠G = 100 what is x?
4. If m∠G = x – 10 and m∠H = 135 what is x?
5. If m∠E = 2x and m∠F = 46 what is x?
6. If m∠F = 3x and m∠E = 39 what is x?
7. If m∠E = 2x + 5 and m∠F = 44 what is x?
8. If m∠F = 2x – 6 and m∠E = 56 what is x?
9. If m∠H = 2(x + 4) and m∠G = 116 what is x?
10.If m∠G = 2(x –5) and m∠H = 120 what is x?
Set C.
ABCD is an isosceles trapezoid.
D C
A B
1. If m∠A = 2x + 10 and m∠B = 3x – 20 what is m∠A?
2. If m∠A = 2x + 15 and m∠B = 4x – 11 what is m∠B?
3. If m∠D = x + 15m and m∠C = 2x –85 what is m∠D?
4. If m∠C = 3y + 12 and m∠D = 2y + 50, what is m∠C?
5. If m∠C = 4x + 70 and m∠D = 2x + 90, what is x?
6. If m∠D = 4x – 20 and m∠C = 5x – 50, what is x?
7. If AC = 60 cm, what is BD?
8. If BD = 70 cm, what is AC?
9. If AC = 4x – 6 and BD = 2x + 10, what is AC?
10.If AC = 3y + 7 and BD = 6y – 8, what is BD?
Lesson 3
Properties of a Parallelogram
A parallelogram is a quadrilateral in which both pairs of opposite sides are
parallel.

13
Do the following:
1. On a graphing paper, draw a parallelogram similar to the one below.
Name your parallelogram ABCD.
2. Draw diagonal AC. What do you notice?
3. Using a protractor, find the measures of the opposite angles of
parallelogram ABCD. Are the angles congruent?
4. Using a protractor, find the measures of each pair of non-opposite
angles. Add their measures. Are the angles supplementary?
5. Using a ruler, find the lengths of each pair of opposite sides. Are their
lengths equal?
6. Draw diagonal BD. What do you notice?
Were you careful in doing the above activity? You actually proved inductively the
following properties of a parallelogram.
1. Each diagonal divides a parallelogram into two congruent triangles.
2. The opposite angles of a parallelogram are congruent.
3. The non-opposite angles of a parallelogram are supplementary.
4. The opposite sides of a parallelogram are congruent.
5. The diagonals of a parallelogram bisect each other.
These properties of a parallelogram can also be proven deductively.
Given: Parallelogram MATH with diagonal MT. H T
Prove: ∆HTM ≅ ∆AMT
M A
D C
A B

14
Proof:
Statements Reasons
1. Parallelogram MATH with
diagonal MT
2. MH // TA
3. ∠HMT ≅ ∠ATM (A)
4. MT ≅ MT (S)
5. HT // MA
6. ∠HTM ≅ ∠AMT (A)
7. ∆HTM ≅ ∆AMT
1. Given
2. A parallelogram is a
quadrilateral in which both pairs
of opposite sides are parallel.
3. If two parallel lines are cut by a
transversal, then any pair of
alternate interior angles are
congruent.
4. Reflexive Property of
Congruence
of opposite sides are
parallel.(Same as reason # 2)
transversal, then any pair of
alternate interior angles are
congruent. ( Same as reason
#3)
7. ASA Congruence
2. The opposite angles of a parallelogram are congruent
H T
Given: Parallelogram MATH
Prove: ∠H ≅ ∠ A
Proof: M A
Statements Reasons
1. Parallelogram MATH
2. Draw MT
3. ∆MHT ≅ ∆TAM
4. ∠H ≅ ∠A
1. Given
2. Two points determine a line.
3. Each diagonal divides a
parallelogram into two congruent
triangles. (First property)
4. Corresponding parts of
congruent triangles are
congruent. (CPCTC)

15
If you want to prove that ∠M ≅ ∠ T, draw diagonal HA. Then follow the above
steps.
3. The non-opposite angles of a parallelogram are supplementary
Given: Parallelogram MATH H T
Prove: ∠H and ∠ M are supplementary
M A
Proof:
Statements Reasons
2. HT // MA
3. ∠H and ∠ M are supplementary
1. Given
of opposite sides are parallel
transversal, then the interior
angles on the same side of the
transversal are supplementary.
4. The opposite sides of a parallelogram are congruent
Given : Parallelogram MATH H T
Prove: HT ≅ MA
HM ≅ TA
M A
Proof:
Statements Reasons
2. Draw diagonal MT
3. ∆MHT ≅ ∆TAM
4. HT ≅ MA
HM ≅ TA
1. Given
2. Two points determine a line
3. Each diagonal divides a
parallelogram into two congruent
triangles.
congruent.

16
5. The diagonals of a parallelogram bisect each other.
Given: Parallelogram LOVE E V
A 2
Prove: EO and LV bisect each other 3 4
1
L O
Proof:
Statements Reasons
1. Parallelogram LOVE with
diagonals EO and LV
2. LE ≅ VO (S)
3. LE // VO
4. ∠1 ≅ ∠2 (A)
5. ∠3 ≅ ∠4 (A)
6. ∆LEA ≅ ∆VOA
7. EA ≅ OA
LA ≅ VA
8. EO and LV bisect each other
1. Given
2. Opposite sides of a
parallelogram are congruent
of opposite sides are parallel
transversal, then the alternate
interior angles are congruent.
5. Vertical angles re congruent
6. SAA Congruence
congruent
8. The bisector of a segment is a
point, line, segment, or plane
that divides the segment into
two segments (Definition of
bisector of a segment)
Example 1
JACK is a parallelogram. If m∠K = 110, what is m∠A.
K C
J A
Solution:
Step 1. Opposite angles of a parallelogram are congruent
∠A ≅ ∠K
m ∠A = m∠K (Congruent angles have equal measures)

17
Step 2. Replace m∠K with 110
m∠A = 110
Example 2 ABCD is a parallelogram. If m∠ A = x + 15 and m∠C = 40,
what is x?
D C
A B
Solution:
Step 1 Opposite angles of a parallelogram are congruent
∠A ≅ ∠C
m∠A = m∠C ( Congruent angles have equal measures)
Step 2. Replace m∠A with x + 15 and m∠C with 40 and solve for x.
x + 15 = 40
x = 40 – 15
x = 25
Example 3
The figure below is a parallelogram. If m∠O = 2x + 10 and m∠E = x +
30, what is m∠O?
E V
Solution: L O
Step 1. Opposite angles of a parallelogram are congruent.
∠O ≅ ∠E
m∠O = m∠E (Congruent angles have equal measures)
Step 2. Substitute 2x +10 for m∠O and x +30 for m∠E. Then solve for x.
2x + 10 = x + 30
2x – x = 30 – 10
x = 20
Step 3. Substitute 20 for x in m∠O = 2x + 10 to solve for m∠O.
m∠O = 2(20) + 10
m∠O = 40 + 10
m∠O = 50

18
Example 4
Quadrilateral ABCD is a parallelogram. If m∠A = 60, what is m∠B?
D C
A B
Solution:
Step 1. Non-opposite angles of a parallelogram are supplementary
∠A and ∠B are supplementary
m∠A + m∠B = 180
Step 2. Substitute 60 for m∠A and solve for m∠B
60 + m∠B = 180
m∠B = 180 – 60
m∠B = 120
Example 5
Quadrilateral ETNA is a parallelogram. If m∠E = x – 60 and m∠A = 2x,
what is m∠E? A N
E T
Solution:
Step 1. Non-opposite angles of a parallelogram are supplementary.
∠E and ∠A are supplementary
m∠E + m∠A = 180
Step 2. Substitute x – 60 for m∠E, and 2x for m∠A.
x –60 + 2x = 180
x + 2x = 180 + 60
3x = 240
x = 80
Step 3. Substitute 80 for x in m∠E = x – 60. Then solve for m∠E.
m∠E = x – 60
m∠E = 80 – 60
m∠E = 20

19
Example 6
In the parallelogram below, m∠A = 2x and m∠C = 4x – 80. What is
m∠B?
D C
A B
Solution:
Step 1. Opposite angles of a parallelogram are congruent.
∠A ≅ ∠C
m∠A = m∠C (Congruent angles have equal measures)
Step 2. Substitute 2x for m∠A and 4x – 80 for m∠C. Then solve for x.
2x = 4x – 80
2x – 4x = – 80
–2x = – 80
x = 40
Step 3. Substitute 40 for x
m∠A = 2x
m∠A =2(40)
=80
Step 4. Non-opposite angles of a parallelogram are supplementary
∠A and ∠B are supplementary
m∠A + m∠B = 180
Step 5. Substitute 80 for m∠A and solve for m∠B.
80 + m∠B = 180
m∠B = 180 – 80
m∠B = 100
Example 7
Quadrilateral EFGH is a parallelogram. If EH = 14 cm long, how long is
side FG?
H G
E F

20
Solution:
Step 1. Opposite sides of a parallelogram are congruent
FG ≅ EH
FG = EH (Congruent segments have equal lengths)
Step 2. Substitute 14 for EH in the equation FG = EH
FG = 14
Therefore FG is 14 cm long.
Example 8
Quadrilateral GEOM is a parallelogram. If MO = 2x + 3 and GE = 4x - 15.
What is MO?
M O
G E
Solution:
Step 1. Opposite sides of a parallelogram are congruent
MO ≅ GE
MO = GE (Congruent segments have equal lengths)
Step 2. Substitute 2x + 3 for MO and 4x – 15 for GE. Solve for x
2x + 3 = 4x – 15
2x – 4x = –15 – 3
–2x = –18
x = 9
Step 3. Substitute 9 for x in MO = 2x + 3
MO = 2(9) + 3
= 18 +3
= 21

21
Example 9
CDEF is a parallelogram. If FD = 12 cm, what is the length of FG?
F E
G
C D
Solution:
The diagonals of a parallelogram bisect each other.
CE bisects FD
Therefore: FG =
2
1
FD
=
2
1
(12)
=6 cm
Example 10
CDEF is a parallelogram with diagonals CE and DF intersecting at G. If
FG = 3x –7 and DG = x +21, find FG.
Solution:
Step 1. Draw the figure.
F E
G
C D
Step 2. The diagonals of a parallelogram bisect each other.
FG = DG
Step 3. Substitute 3x – 7 for FG and x + 21 for DG in the equation FG DG.
3x – 7 = x + 21
3x –x = 21 + 7
2x = 28
x = 14

22
Step 4. Substitute 14 for x in the equation FG = 3x – 7 . (See given data)
FG = 3x – 7
=3(14) – 7
=42 – 7
= 35
Try this out
Set A
Fill in the blanks
Quadrilateral ABCD is a parallelogram..
D C
500
880
A B
1. m∠A = ________
2. m∠ABC = _____
3. m∠A + m∠ABC = ________
4. m∠ABC + m∠C = ________
5. m∠C + m∠ADC = ________
6. ∆ ABD ≅ ________
Quadrilateral EFGH is a parallelogram.
H G
I
E F
7. HI = __________
8. EI = ________
9. If EG is 16 cm, then EI = ________
10.If HI is 7 cm, then FI = __________

23
Set B
Quadrilateral ABCD is a parallelogram
D C
A B
1. If m∠A is 50 , what is m∠C?
2. If m∠B is 125 , what is m∠D?
3. If m∠A = x and m∠C = 30, what is x?
4. If m∠B = y and m∠D = 115, what is y?
5. If m∠A = x + 30, and m∠C = 60, what is x?
6. If m∠D = 100 and m∠B = y – 40, what is y?
7. If m∠A = 50, what is m∠D?
8. If m∠D = 130, what is m∠C?
9. If m∠B = 2x – 20 and m∠D = x + 40, what is m∠B?
10.If m∠A = 2x – 50 and m∠C = x + 10 what is m∠A?
Set C
Use the figure below for exercises 1-10
Quadrilateral ABCD is a parallelogram.
D C
E
A B
1. If AB = 17 cm, then CD = _________
2. If AD = 25 cm, then BC = __________
3. If AE = 5 dm, then CE = __________
4. If BD = 36 cm, then BE = __________
5. If AB = 2x + 10 and CD = 15, then AB = _________
6. If AD = 4x + 15 and BC = 2x + 21 then AD = _________
7. If AB = x + 6 and CD = 14, what is x?
8. If AD = 20 and BC = x – 5, what is x?
9. If AE = 15 and CE = x + 4, what is x?
10.If BE = 2x and DE = 6, what is x?

24
Let’s summarize
1. A trapezoid is a quadrilateral with exactly one pair of parallel sides.
2. The median of a trapezoid is a segment joining the midpoints of the non-
parallel sides of a trapezoid.
3. The median of a trapezoid is parallel to its bases and half the sum of the
lengths of the bases.
4. An isosceles trapezoid is a trapezoid with congruent non-parallel sides.
6. The diagonals of an isosceles trapezoid are congruent.
7. A parallelogram is a quadrilateral in which both pairs of opposite sides are
parallel.
9. Opposite angles of a parallelogram are congruent.
10.Non-opposite angles of a parallelogram are supplementary.
11.The diagonals of a parallelogram bisect each other.
Multiple Choice. Choose the letter of the correct answer.
1. Non-opposite angles of a parallelogram are
A. Complementary C. Adjacent
B. Supplementary D. Congruent
C.
2. A quadrilateral with exactly one pair of parallel sides
A. Square C. Trapezoid
B. Rectangle D. Rhombus
3. In the figure at the right, DC = 20 cm
And AB = 36 cm. What is EF? D C
A. 16 cm
B. 56 cm E F
C. 28 cm
D. 46 cm A B
4. The figure below is a parallelogram. If AD = 2x - 10
and BC = x + 30, then BC =
A. 50 D C
B. 60
C. 70
D. 80 A B

25
5. The figure below is a rhombus. If m ∠I = 4x and m ∠E = 2x + 60, what is
m ∠I in degrees? E V
A. 100
E. 110
F. 120
G. 130 G I
6. Quadrilateral BEST is a parallelogram. If m ∠B = x + 40 and
m ∠E = 2x + 20, what is m ∠B in degrees?
A. 50 T S
B. 60
C. 70
D. 80
B E
7. The figure below is a parallelogram. The diagonals AC and BD intersect at
E. If AE = 2x and EC = 12, what is x?
A. 5 D C
B. 6
C. 7 E
D. 9
A B
8. Quadrilateral CDEF is a parallelogram. If m ∠C = y and m ∠E = 2y – 40,
then m ∠D is
A. 80 F E
B. 110
C. 140
D. 170
C D
9. Into how many congruent triangles is a parallelogram divided by one of its
diagonals?
A. 1 C. 3
B. 2 D. 4
10.Base angles of an isosceles trapezoid are
A. Complementary C. Congruent
B Supplementary D. Adjacent

26
Answer Key
1. False
2. True
3. False
4. False
5. 17.5
6. AB = 16
7. BE + TS = 30
8. m∠ADC = 130
9. AC = 12
10.m∠C = 40
Try this out
Lesson 1
Set A
1. 7
2. 9.5
3. 11.5
4. 8.5
5. 14
6. 18
7. 6.2
8. 9.6
9. 14
10.24
Lesson 2
Set A
1. ∠P and ∠Q; ∠R and ∠S
2. AD and BC
3. DC and AB
4. m∠B = 30
5. m∠A = 60
6. m∠D = 110
7. m∠C = 105
8. m∠D = 145
9. m∠B = 60
10.m∠C = 145
Set B
1. x = 10
2. y = 17
3. x = 10
4. y = 13
5. x = 10
6. y = 18
7. x = 9
8. y = 11
9. x = 2
10.x = 14
Set C
1. x = 15.5
2. x = 23
3. NM = 20
4. KL = 29
5. x = 16
6. x = 25
7. OP = 23
8. NM = 22
9. OP = 15
10.NM = 10
Set B
1. x = 40
2. x = 85
3. x = 80
4. x = 145
5. x = 23
6. x = 13
7. x = 19.5
8. x = 31
9. x = 54
10.x = 35
Set C
1. m∠A = 70
2. m∠B = 41
3. m∠D = 115
4. m∠C = 126
5. x = 10
6. x = 30
7. BD = 60
8. AC = 70
9. AC = 26
10.BD = 22

27
Lesson 3
Set A
1. 50
2. 130
3. 180
4. 180
5. 180
6. 42
7. FI
8. GI
9. 8 cm
10.7 cm
1. B
2. C
3. C
4. C
5. C
6. D
7. B
8. C
9. B
10.C
Set B
1. m∠C = 50
2. m∠D = 125
3. x = 30
4. y = 115
5. x = 30
6. y = 140
7. m∠D = 130
8. m∠C = 50
9. m∠B = 100
10.m∠A = 70
Set C
1. CD = 17 cm
2. BC = 25 cm
3. CE = 5 dm
4. BE = 18 cm
5. AB = 15
6. AD = 27
7. x = 8
8. x = 25
9. x =11
10.x = 3

Module 2
Properties of Quadrilaterals
This module is about the properties of the diagonals of special quadrilaterals. The
special quadrilaterals are rectangles, square, and rhombus. The conditions sufficient to
guarantee that a quadrilateral is a parallelogram are also discussed in this module.
This module is designed for you to
1. apply inductive/deductive skills to derive the properties of the diagonals of special
quadrilaterals
• rectangle
• square
• rhombus
2. verify sets of sufficient conditions which guarantee that a quadrilateral is a
parallelogram
3. apply the conditions to prove that a quadrilateral is a parallelogram
4. solve routine and non routine problems
True of False
1. The diagonals of a square are congruent.
2. The diagonals of a rectangle are perpendicular.
3. The diagonals of a rhombus bisect each other.
4. A square is a rhombus.
5. If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral
is a parallelogram.

2
6. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a
parallelogram.
Quadrilateral ABCD is a rectangle. Its diagonals AC and BD intersect at E.
D C
E
A B
7. If AC = 2(x + 10) and BD = x + 60, what is AC?
8. If AE = 4x – 5 and CE = 10 + x, what is AE?
9. Quadrilateral CDEF is a rhombus. If m∠FCE = 3x - 5 and m∠DCE = 2x, find
m∠FCD.
F E
C D
Quadrilateral GHIJ is a square.
J I
G H
10.If m∠HGI is 3(x + 5), what is x?

3
What you will do
Lesson 1
The Properties of the Diagonals of Special Quadrilaterals
A diagonal of a quadrilateral is a segment which connects any two non-consecutive
vertices. In the following quadrilateral, AC and BD are the diagonals.
D C
A B
The following are the properties of the diagonals of special quadrilaterals.
1. The diagonals of a rectangle are congruent.
3. The diagonals of a square are perpendicular
4. Each diagonal of a square bisects a pair of opposite angles.
5. The diagonals of a rhombus are perpendicular.
6. Each diagonal of a rhombus bisects a pair of opposite angles
You can apply inductive skills to derive these properties of the diagonals of special
quadrilaterals. In the following activities you need a ruler, a pencil, a protractor and pieces
of graphing paper.
1. Do the following activity:
a. On a graphing paper, draw a rectangle.
b. Name your rectangle ABCD.
c. Draw diagonals AC and BD.
d. Find the lengths of AC and BD. Are their lengths equal? Are the diagonals
congruent?
Conclusion: The diagonals of a rectangle are congruent
a. On a graphing paper, draw a square.
b. Name your square ABCD.
c. Draw diagonals AC and BD.

4
d. Find the lengths of the diagonals. Are their lengths equal? Are the diagonals of
the square congruent?
Conclusion: The diagonals of a square are congruent.
a. Construct a square on a graphing paper
b. Name your square EFGH.
c. Draw its diagonals EG and HF.
d. Label the intersection of the diagonals, M.
e. Using a protractor, find the measures of ∠HME, and ∠HMG.
f. What kind of angles are the two angles?
g. Are the diagonals perpendicular?
Conclusion: The diagonals of a square are perpendicular
4. Do the following activity.
a. Draw a square on a graphing paper.
b. Name your square ABCD.
c. Draw diagonal AC.
d. What do you notice? Into how many angles are the two
opposite vertex angles divided?
e. What do you conclude?
Conclusion: Each diagonal of a square bisects a pair of opposite angles.
a. Draw a rhombus on a graphing paper.
b. Name your rhombus ABCD.
c. Draw the diagonals and name the point of intersection, E.
d. Find the measures of ∠AED and ∠CED.
e. What kind of angles are they?
f. What can you say about the diagonals?
Conclusion: The diagonals of a rhombus are perpendicular.
a. Draw a rhombus on a graphing paper.
b. Name your rhombus ABCD.
c. Draw diagonal AC.
d. What do you notice? Into how many angles are the two
opposite vertex angles divided?
e. What do you conclude?
Conclusion: Each diagonal of a rhombus bisects a pair of opposite angles.

5
These properties of the diagonals of special quadrilaterals can also be proven
deductively. Let us prove the first three properties deductively.
1. The diagonals of a rectangle are congruent. D C
Given: Rectangle ABCD
with diagonals AC and BD.
Prove: BD ≅ AC
A B
Proof:
Statements Reasons
1. Rectangle ABCD with diagonals
AC and BD
2. AD ≅ BC (S)
3. ∠DAB and ∠CBA are right
angles
4. ∠DAB ≅ ∠CBA (A)
5. AB ≅ AB (S)
6. ∆DAB ≅ ∆ CBA
7. BD ≅ AC
1. Given
2. Opposite sides of a
parallelogram are congruent
(Remember, a rectangle is a
parallelogram)
3. A rectangle is a parallelogram
with four right angles
4. Any two right angles are
congruent
5. Reflexive Property of
Congruence
6. SAS Congruence
7. Corresponding Parts of
Congruent Triangles are
Congruent
Triangles ∆DAB and ∆ CBA overlap. If you find difficulty visualizing the two
overlapping triangles, separate the figure into two triangles .
D C D C
A B A B A B
2. The diagonals of a square are congruent. D C
Given: ABCD is a square with
diagonals AC and BD
Prove: BD ≅ AC A B

6
Proof
Statements Reasons
1. ABCD is a square with
diagonals AC and BD
2. AD ≅ BC (S)
3. ∠DAB and ∠CBA
are right angles
4. ∠DAB ≅ ∠CBA (A)
5. AB ≅ AB (S)
6. ∆ DAB ≅ ∆ CBA
7. BD ≅ AC
1. Given
2. Opposite sides of a parallelogram
are congruent
(Remember, a square is a
parallelogram)
3. A rectangle has four right angles
(Remember that a square is a
rectangle with four congruent sides
and a rectangle has four right
angles.)
4. Any two right angles are congruent
5. Reflexive Property of Congruence
6. SAS Congruence
7. Corresponding Parts of Congruent
Triangles are Congruent
Given: TEAM is a square with diagonals AT and ME M A
Prove: ME ⊥ AT
O
T E
A proof can also be written in paragraph form.
Proof:
Side TM and side EA are congruent since they are sides of a square. A square is a
rectangle with four congruent sides. MO ≅ MO by Reflexive Property of Congruence. The
diagonals of a parallelogram bisect each other. Since a square is a parallelogram therefore
TO ≅ AO. ∆MOT ≅ ∆MOA by SSS congruence. Since ∠MOT and ∠MOA are supplementary
and congruent, then each of them is a right angle. Therefore ME ⊥ AT by the definition of
perpendicular.

7
Example 1
The figure at the right is a rectangle. E V
If the diagonal LV = 2x and the diagonal
OE = 12 cm, find x.
Solution:
Step 1. The diagonals of a rectangle are congruent. L O
LV ≅ OE
LV = OE (Congruent segments have equal lengths)
Step 2. Substitute 2x for LV and 12 for OE. Then solve for x.
2x = 12
x = 6
Answer: The value of x is 6 cm.
Example 2
D C
Quadrilateral ABCD at the right is a square.
Find m∠CAB
Solution:
Step 1, Quadrilateral ABCD is a square and
a square is a rectangle.
A B
Therefore: m∠DAB = 90.
Step 2. But each diagonal of a square bisects a pair of opposite angles.
Hence: m ∠CAB =
2
1
m∠DAB
Step 3. Substitute 90 for m∠DAB.
m ∠CAB =
2
1
(90)
= 45
Answer: m ∠CAB =45
A N
Example 3 EDNA is a square.
If m∠END is 3(x +5), what is x?
E D

8
Solution:
a. m∠DCB = 90 since ABCD is a square
b. Each diagonal of a square bisects a pair of
opposite angles.
Hence: m∠ACB = 45
3(x + 5) = 45
3x + 15 = 45
3x = 45 – 15
3x = 30
x = 10
D C
Example 4
The figure at the right is a rhombus.
If m ∠CAB = 30 , what is the m ∠ CAD?
Step 1. Each diagonal of a rhombus bisects
pair of opposite angles.
A B
m ∠ CAD = m ∠CAB
Step 2. Substitute 30 for m ∠CAB in the above equation.
m ∠ CAD = 30
Answer: The measure of ∠ CAD is 300
.
Example 5
DEFG is a rhombus. If m∠FGE = 5x – 8 and m∠DGE = 3x + 22, find the measure
of (a) m ∠FGE (b) m∠DGE and (c) m∠FGD
G F
D E
Solution:
Step 1. Each diagonal of a rhombus bisects a pair of opposite angles.
m∠FGE = m∠DGE
5x – 8 = 3x + 22
5x – 3x = 22 + 8
2x = 30
x = 15

9
Step 2. Substitute 15 for x
a. m∠FGE = 5x – 8
= 5(15) – 8
= 67
b. m∠DGE = 3x + 22
=3(15) + 22
= 67
c. m∠FGD = m∠FGE + m∠DGE
= 67 + 67
= 134
Answers: (a) m ∠FGE = 67
(b) m∠DGE = 67
(c) m∠FGD = 134
Example 6
BETH is a rhombus. If m∠TBE = 35, H T
what is m∠HEB?
M
Solution:
Step 1. The diagonals of a rhombus B E
are perpendicular. Hence, ∠BME
is a right angle and its measure is 900
.
m∠BME = 90
Step 2. The sum of the measures of the angles of a triangle is 1800
m∠TBE + m∠BME+ m∠HEB = 180
Step 3. Substitute 35 for m∠TBE and 90 for m∠BME in the above equation.
35 + 90 + m∠HEB = 180
125 + m∠HEB = 180
m∠HEB = 180 – 125
m∠HEB = 55.
Answer: m∠HEB = 55
D C
Example 7
M
ABCD is a rhombus. If AM = 16 cm,
what is CM?
A B

10
Solution:
Step 1. The diagonals of a rhombus
Bisect each other.
CM = AM
Step 2. Substitute 16 cm for AM in the above equation.
CM = 16 cm
Answer: CM = 16 cm
Try this out
Set A . ABCD is a rectangle.
D C
A B
True or False
1. The lengths of AC and BD are equal.
2. Diagonals AC and BD are perpendicular.
3. The diagonal AC bisects ∠DCB.
4. A rectangle is a parallelogram.
ABCD is a square
D C
E
5. ∠EAB ≅ ∠EBC
6. ∠DEC is a right angle.
A B
FGHI is a rhombus.
I H
7. m∠FIG = m∠HIG J
8. The sum of m∠JFG and m∠JGF is 45.
9. ∆FIG ≅∆ HGI
F G
10.The diagonal FH bisects the rhombus into two congruent triangles.

11
Set B
ABCD is a rectangle
D C
A B
Find the indicated measure
1. AC = 15 dm. Find BD
2. BD = 23 cm. Find AC
EFGH is a rhombus
Find the indicated measure. H G
3. m∠HGE = 35. Find m∠FGE. I
4. m∠HEI = 20. Find m∠FEI.
5. m∠IEF = 30. Find m∠IFE
6. m∠IHE = 58. Find m∠IEH
7. m∠IEF = 20. Find m∠IEF + m∠EIF E F
8. m∠IGH = 25. Find m∠IGH + m∠HIG
9. If ABCD is a square, D C
then m∠ACB = ________
10.If ABCD is square,
then m∠ DEC =___________ E
A B
Set C
ABCD is a rectangle with diagonals AC and BD. D C
1. AC = 2x + 15, BD = 3x + 10. Find AC.
2. BD = 6x + 5, AC = 5x + 14. Find BD.
A B

12
FERM is a rhombus.
M R
I
F E
3. If m∠IFE = x +20, m∠IEF = x + 26 ,find x.
4. If m∠IMR = 4x + 20, m∠IRM = 2x + 10, find x.
5. m∠IFE + m∠IEF = __________
6. m∠IMR + m∠IRM = _________
BETH is a square.
H T
B E
7. If HM = x + 15, HE = 40, what is x?
8. If EM = x + 9, HE = 30, what is x?
9. If BM = x + 12, EM = 2x – 20, what is x?
10.If HM = 44 – x, TM = 4 + 3x, what is x?
Lesson 2
Conditions for a Parallelogram
The following are some conditions which guarantee that a given quadrilateral is a
parallelogram.
1. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral
is a parallelogram.
is a parallelogram.
3. If one pair of opposite sides of a quadrilateral are both congruent and parallel, then
the quadrilateral is a parallelogram.
parallelogram.

13
5. If the non-opposite angles of a quadrilateral are supplementary, then the quadrilateral
is a parallelogram.
You can verify these sets of sufficient conditions which guarantee that a
quadrilateral is a parallelogram. In the following activities you need a pencil, a ruler, a
protractor and pieces of bond paper and graphing paper.
is a parallelogram.
Do this activity:
a. On a graphing paper, draw a quadrilateral such that both pairs of opposite sides
are congruent. ( See the illustration.)
b. Are the opposite sides equidistant? Find this out by using a ruler.
c. Are both pairs of sides parallel? (Remember, parallel lines are everywhere
equidistant.)
d. Can you now conclude that the quadrilateral is a parallelogram? Why?
is a parallelogram.
a. On a graphing paper, with the aids of a ruler and a protractor, construct an
quadrilateral such that both pairs of opposite angles are congruent. (See
illustration)

14
b. Are the opposite sides congruent?
c. Can you now conclude that the quadrilateral is a parallelogram? Why?
3. If one pair of opposite sides of a quadrilateral are both congruent and parallel, then
Do this activity
a. On a graphing paper, draw a quadrilateral such that one pair of opposite sides
are both congruent and parallel.
( See illustration below)
b. Are the other two opposite sides congruent?
c. Can you now conclude that the quadrilateral is a parallelogram? Why?
parallelogram.

15
a. On a bond paper, draw segments AC and BD bisecting each other. (See the
illustration below.)
D C
A B
b. Connect A to B, B to C, C to D and D to A .
D C
A B
c. Using a ruler, find the lengths of AB and CD. Are they equal?
d. Using a ruler, find the lengths of AD and BC. Are the lengths equal?
e. What kind of quadrilateral is ABCD?
5. If the non-opposite angles of a quadrilateral are supplementary, then the quadrilateral
is a parallelogram.
a. On a bond paper, draw angle A. (See the illustration below.)
A
b. Draw angle ADC such that its measure is supplementary to that of angle A.
● C
A
D

16
c. Draw angle DCB such that its measure is equal to that of angle A.
B
● C
A D
d. Find the measure of angle CBA. Is it equal to the measure of angle ADC? Are
∠A and ∠B supplementary? How about ∠B and ∠C? How about ∠D and ∠C?
Why/
e.What kind of quadrilateral is ABCD?
Example 1
Determine whether the figure is a parallelogram. Identical “tick marks” indicate that
the sides are congruent and identical “arrowheads” indicate the lines are parallel.
Solution:
If one pair of opposite sides of a quadrilateral are both congruent and parallel, then the
quadrilateral is a parallelogram.
Hence the geometric figure is a parallelogram.
Example 2
Determine whether the figure is a parallelogram.
Solution:

17
A pair of alternate interior angles are congruent, therefore a pair of opposite sides are
parallel. These parallel sides are also congruent. As can be seen in the figure, they have the
same length.
Hence the figure is a parallelogram.
Example 3.
Find the value of x for which ABCD is a parallelogram.
A D
270
3x
B C
Solution:
If two lines are cut by a transversal and a pair of alternate interior angles are
congruent, then the lines are parallel.
AD // BC since ∠ADB ≅∠CBD
CD // AB if 3x = 27
x = 9
Hence the value of x should be 9.
Try this out
Set A
True or False
is a parallelogram.
is a parallelogram.
3. If one pair of opposite sides of a quadrilateral are parallel, then the quadrilateral is a
parallelogram.
parallelogram.
5. If the opposite angles of a quadrilateral are supplementary , then the quadrilateral is
a parallelogram.

18
ABCD is a quadrilateral. AD = 5 cm and AB = 9 cm.
y
D C
5 cm
x
A B
9 cm
6. ABCD is a parallelogram if x = 5 cm and y = 9 cm.
7. ABCD is a parallelogram if m∠C = 60 and m∠B = 120.
8. ABCD is a parallelogram if AB // DC.
9. ABCD is a parallelogram if m∠B ≅m∠D
10. ABCD is a parallelogram if AB ≅ DC ≅ AD ≅ BC.
Set B.
Determine whether each quadrilateral is a parallelogram. Identical “tick marks”
indicate that the sides or angles are congruent and identical “arrowheads” indicate the lines
are parallel.
D C
1.
A B
D C
2.
A B
3.

19
4.
5.
6.
10
7.
6 7
10
8.
1000
790
810
1000
300
9. 500
500
300

20
15 cm
10. 300
300
15 cm
Set C.
What values of x and y guarantee that each quadrilateral is a parallelogram.
1. 6.
y 500
450
1350
x y
3x y
2y
2 1100
700
7.
x 8 cm
x y
14 cm
y 126
3. 8.
6 cm
x 90 x
12 cm 3y
15 cm 2x + 10
4. 9.
y 4 cm 2y 24
70
x
4y
2x 600
5. 10, 5
2x – 5
y 1200
32

21
Let’s summarize
1. A diagonal of a quadrilateral is a segment which connects any two non-consecutive
vertices.
2. The diagonals of a rectangle are congruent.
5. Each diagonal of a square bisects a pair of opposite angles.
6. The diagonals of a rhombus are perpendicular.
7. Each diagonal of a rhombus bisects a pair of opposite angles
8. A square is a special type of rhombus.
is a parallelogram.
10.If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral
is a parallelogram.
11.If one pair of opposite sides of a quadrilateral are both congruent and parallel, then
12.If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a
parallelogram.
13.If the non-opposite angles of a quadrilateral are supplementary, then the quadrilateral
is a parallelogram.
14.A quadrilateral is a parallelogram if both pairs of opposite side are parallel
Multiple Choice. Choose the letter of the correct answer.
1. A parallelogram is a rhombus if
A. The diagonals bisect each other
B. The diagonals are perpendicular.
C. Two consecutive angles are supplementary.
D. The opposite sides are parallel.
2. Which of the following is sufficient to guarantee that a quadrilateral is a
parallelogram?
A. The diagonals are perpendicular
B. A pair of adjacent sides are congruent
C. Two consecutive angles are congruent
D. The diagonals bisect each other

22
3. ABCD is a rectangle. if diagonal AC = 2x + 6 and diagonal BD = 10, what is x?
A. 1 C. 3
B. 2 D. 4
4. ABCD is a rhombus.
D C
A B
If m∠DCA = 2(x+8) and m∠BCA = 3x + 9, what is m∠DCB?
A. 40 C. 60
B. 50 D. 70
5. ABCD is a square.
D C
A B
If m∠ABD = 3(x + 10), what is x?
A. 1 C. 5
B. 3 D. 7
6. ABCD is a rhombus. Diagonals AC and BD intersect each other at E.
D C
E
A B
If AE = 12 and CE = 3x, what is x?
A. 2 C. 6
B. 4 D. 8

23
7. ABCD is a rhombus . Diagonals AC and BD intersect at E.
D C
E
A B
What is m∠AED?
A. 30 C. 60
B. 45 D. 90
8. What values of x and y guarantee that ABCD is a parallelogram.
D C
y 64
x y
A. x = 64 , y = 116 C. x = 64, y = 64
B. x = 32, y = 116 D. x = 32, y = 64
9. Find the value of x for which ABCD is a parallelogram.
D C
400
800
800
2x
A B
A. 10 C. 30
B. 20 D. 40
10. Find the value of x for which ABCD is a parallelogram.
18
3x – 6 12
18
A. 8 C. 4
B. 6 D. 2

24
Answer Key
1. True
2. False
3. True
4. True
5. True
6. True
7. AC = 100
8. AE = 15
9. m∠FCD = 20
10.x = 10
Lesson 1
Set A
1. True
2. False
3. False
4. True
5. True
6. True
7. True
8. False
9. True
10.True
Set B
1. 15
2. 23
3. 35
4. 20
5. 60
6. 32
7. 110
8. 115
9. 45
10.90
Set C
1. AC = 25
2. BD = 59
3. x = 22
4. x = 10
5. 90
6. 90
7. x = 5
8. x = 6
9. x = 32
10.x = 10
Lesson 2
Set A
1. True
2. True
3. False
4. True
5. False
6. True
7. True
8. True
9. True
10.False
Set B
1. Parallelogram
2. Parallelogram
3. Parallelogram
4. Parallelogram
5. Parallelogram
6. Parallelogram
7. Not a parallelogram
8. Not a parallelogram
9. Parallelogram
10.Parallelogram
Set C
1. x = 500
y = 1300
2. x = 700
y = 1100
3. x = 6 cm
y = 12 cm
4. x = 15 cm
y = 4 cm
5. x = 600
y = 600
6. x = 450
y = 450
7. x = 8 cm
y = 7 cm
8. x = 90 units
y = 42 units
9. x = 30 units
y = 12 units
10.x = 5 units
y = 8 units
1. B
2. D
3. B
4. C
5. C
6. B
7. D
8. A
9. B
10.B

MODULE 1
Geometric Relations
This module is about relations of segments and angles. As you go over the
exercises, you will develop your skills involving points, segments and angle pairs and solve
problems on the relationships between segments and between angles. Treat the lessons
with fun and take time to go back if you feel you are at loss.
This module is designed for you to:
1. illustrate betweenness and collinearity.
2. illustrate the following:
• congruent segments
• midpoint of a segment
• congruent angles
• bisector of an angle
• complementary angles
• supplementary angles
• adjacent angles
• linear pair
• vertical angles

2
Answer the following questions asked from the given figure:
. A D
I M H J P
. . . . . . . . . .
-3 E O 3 6
F C
Answer the following questions:
1. What is /IP/?
2. If /IM/ = /MH/ = /IH/, then what point is between the other two?
3. If M is the midpoint of AC, what segments are congruent?
4. Name the coordinate of the midpoint of IJ
5. ∠ EMF and ∠ FMC are ____ angles.
6. ∠ EMF form a linear pair with ______.
7. If ∠ EMF is the complement of ∠ FMC and on ∠ FMC = 75, what is on ∠ EMF?
8. If ∠ AMD ≅ ∠ DMC and are supplementary, what kind of angle is each?
9. What is the measure of each angle if the measure of the angles in a
supplementary pair is twice that of the other?
10. What is the measure of each angle if the two angles are both vertical and
complementary?

3
What you will do
Lesson 1
Collinearity of Points
You may ask this question, “how many points are there in a line?” “ How many
points are there in a plane?” Do you know the answer? Yes, it is infinite or many points.
Now you take a look at BP in the figure.
• .X
B C P
• Y
•Z
Points B, C, P are contained in BP
Points X, Y, Z are not in BP
Points B, X, Y are not in BP
The set of points B, C, P are collinear
The set of points X, Y, Z are not collinear
The set of points B, X, Y are not collinear
Collinear points is a set of points which are contained in a line
Examples:
The following set of points are collinear
1. A, O, P, L S
2. M, N, P, S A O P L
M
N

4
A
3. A, Q, C
P . . Q
4. A, P, B, S
5. B, R, C . B .R . C
S
Do you still remember the number line?
A B C D E F G H I J K L M N
. . . . . . . . . . . . . . .
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
The points of a line can be placed in correspondence with the real numbers in such a
way that:
1. to every point of the line there corresponds exactly one real number;
2. to every real number there corresponds exactly one point of the line;
3. the distance between any two points is the absolute value of the
corresponding numbers.
The number corresponding to a given point is called the coordinate of the point.
Examples:
1. The coordinate of M is 6, the coordinate of G is 0, the coordinate of B is -5.
B C
. . . . .
X -1 O 1 Y
2. If the coordinate of B is x and the coordinate of C is y then, /BC/ = /X-Y/ (read as
distance BC equals the absolute value of X minus Y).

5
J B
. . . . . . . . . . .
-5 O 3
3) /JB/ = │-5 - +3│
= │-8│
= 8
Try this out
Which set of points are collinear:
C
B
.
1. A, O D D E
2. A, B, O A O
3. B, O, F
4. G, O, F
5. C, O G G . F
A B C D E F G H I J K L M
. . . . . . . . . . . . .
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5
Give the coordinate of each of the following points:
6. G
7. D
8. M
9. A
10. K
B. The coordinates of P and Q are listed. Find /PQ/.
1. P: O 6. P: 7
Q: 7 Q: 3

6
2. P: 12 7. P: 19
Q: 0 Q: 113
3. P: 4 8. P: 5
Q: 15 Q: - 5
4. P: 21 9. P: 56
Q: 14 Q: -18
5. P: 15 10. P: -12
Q: 6 Q: -51
M N O P Q R S C T U V W X Y Z A B
C. . . . . . . . . . . . . . . . . . . . . .
-5 -4 -3 -2 -1 0 1 2 3 4 5
-4.5 0.5 ∏
Find:
1. /PS/ 6. /RX/ + /PR/
2. /US/ 7. /CR/ + /OR/
3. /PT/ 8. /PS/ + /RU/
4. /NC/ 9. /SW + /WY/
5. /NT/ 10. /NR/ + /RW/
Lesson 2
Betweenness
A B C
. . .
Let A, B, C be three points. B is between A and C. If A, B and C are on one line and
/AB/ + /BC/ = /AC/. This definition of betweenness means that:
1. If B is between A and C, then A, B and C then, A, B, C are collinear and /AB/ +
/BC/ = /AC/.
2. If A, B and C are collinear and /AB/ + /BC/ = /AC/ then, B is between A and C.

7
Examples:
1. If x, y, and z are collinear X Y Z
and /xy/ + /yz/ = /xz/ then,
Y is between X and Z.
R
2. Given: T is between R and U
Conclusion: R, T, U are collinear
and /RT/ + /TU/ = /RU/ . T
.U
3. O is between S and P S O P
Find /SP/ . . .
Solution: -2 0 3
/SO/ + /OP/ = /SP/
2 + 3 = /SP/
5 = /SP/
Try this out
A. Which point is between the other two?
1.
. P
. O
. M
C E U
2.
B O A
3.
4. A, B, and C are three points on a line with coordinates 8, 4, and 13 respectively.

Grade 9 (Alternate) Mathematics III - Learning Modules for EASE Program of DepEd

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