2. 𝑫𝒆𝒇𝒊𝒏𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝒂 𝑷𝒐𝒊𝒏𝒕 𝒕𝒐 𝒂 𝑳𝒊𝒏𝒆
The distance from a point to a line is the length of the perpendicular
segment from the point to the line. If the point lies on the line, the
distance is zero.
P
C
A B
The distance from point P to AB is PC.
Here PC ⊥ AB
3. Theorem 35
A point on the bisector of an angle is equidistant from the sides of the
angle.
A
D
G
F
E
C
B
Given: 𝑩𝑫 bisects ∠𝑨𝑩𝑪, G lies on 𝑩𝑫
Prove: 𝑮𝑬 = 𝑮𝑭
4. A
D
G
F
E
C
B
Given: 𝑩𝑫 bisects ∠𝑨𝑩𝑪, G lies on 𝑩𝑫
Prove: 𝑮𝑬 = 𝑮𝑭
𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭𝐬 𝐑𝐞𝐚𝐬𝐨𝐧𝐬
1. 𝐁𝐃 bisects ∠𝐀𝐁𝐂,
G lies on 𝐁𝐃
𝐆𝐢𝐯𝐞𝐧
2. ∠𝐀𝐁𝐃 ≅ ∠𝐂𝐁𝐃 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐀𝐧𝐠𝐥𝐞 𝐁𝐢𝐬𝐞𝐜𝐭𝐨𝐫
3. 𝐆𝐄 ⊥ 𝐁𝐀
𝐆𝐅 ⊥ 𝐁𝐂
𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞
𝐟𝐫𝐨𝐦 𝐚 𝐏𝐨𝐢𝐧𝐭 𝐭𝐨 𝐚 𝐋𝐢𝐧𝐞
4. ∠𝐆𝐄𝐁 𝐚𝐧𝐝 ∠𝐆𝐅𝐁 𝐚𝐫𝐞 𝐫𝐢𝐠𝐡𝐭 𝐚𝐧𝐠𝐥𝐞𝐬. 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐏𝐞𝐫𝐩𝐞𝐧𝐝𝐢𝐜𝐮𝐥𝐚𝐫
5. 𝐁𝐆 ≅ 𝐁𝐆 𝐑𝐞𝐟𝐥𝐞𝐱𝐢𝐯𝐞 𝐏𝐫𝐨𝐩𝐞𝐫𝐭𝐲 𝐨𝐟 𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐜𝐞
6. ∆𝐆𝐄𝐁 𝐚𝐧𝐝 ∆𝐆𝐅𝐁 𝐚𝐫𝐞 𝐫𝐢𝐠𝐡𝐭 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬. 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐑𝐢𝐠𝐡𝐭 𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬
7. ∆𝐆𝐄𝐁 ≅ ∆𝐆𝐅𝐁 𝐇𝐲𝐩𝐨𝐭𝐞𝐧𝐮𝐬𝐞 − 𝐀𝐜𝐮𝐭𝐞 𝐀𝐧𝐠𝐥𝐞
8. 𝐆𝐄 ≅ 𝐆𝐅 𝐂𝐏𝐂𝐓𝐂
9. 𝐆𝐄 = 𝐆𝐅 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐭 𝐒𝐞𝐠𝐦𝐞𝐧𝐭𝐬
proof:
5. Theorem 36
A point equidistant from the sides of an angle lies on the bisector of
an angle
A
D
G
F
E
C
B
Given: 𝐺𝐸 ⊥ 𝐵𝐴; 𝐺𝐹 ⊥ 𝐵𝐶, 𝐺𝐸 = 𝐺𝐹
Prove: 𝐺 lies on the bisector of ∠𝐴𝐵𝐶.
𝑷𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓 𝑩𝒊𝒔𝒆𝒄𝒕𝒐𝒓𝒔 𝒐𝒇 𝒂 𝑺𝒆𝒈𝒎𝒆𝒏𝒕
Recall that the perpendicular bisector of a
segment is a line, ray, segment, or plane that is
perpendicular to the segment at its midpoint.
6. 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟑𝟕
A point on the perpendicular bisector of a
segment is equidistant from the endpoint of a segment.
C
P
D
B
A
Given: 𝑪𝑫 ⊥ 𝒃𝒊𝒔𝒆𝒄𝒕𝒐𝒓 𝒐𝒇 𝑨𝑩, 𝑷 lies on 𝑪𝑫.
Prove: 𝑷𝑨 = 𝑷𝑩
8. 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟑𝟖
A point equidistant from the endpoint of a segment lies
on the perpendicular bisector of a segment.
C
P
D
B
A
Given: 𝐏𝐀 = 𝐏𝐁
Prove: P lies on the perpendicular bisector of AB.
10. 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟑𝟗
If a line contains two points each of which is
equidistant from the endpoints of the segment, then
the line is the perpendicular bisector of the segment.
C
P
D B
A
Given: AP=BP, AC=BC
Prove: PD is the ⊥ bisector of AB.
12. 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟒𝟎
In a plane, through a given points on a line, there is
exactly one line perpendicular to the line.
R
S
M P Q
Given: MQ and point P on MQ
Prove: a. There is at least on line PR ⊥ MQ
b. There is only one line PR ⊥ MQ
Part 1: Show that there is one line 𝑷𝑹 ⊥ 𝑴𝑸 through point P.
Through the end point of 𝑷𝑸 , ray 𝑷𝑹 can be constructed in such a way
that 𝒎∠𝑹𝑷𝑸 = 𝟗𝟎. ∠𝑹𝑷𝑸is therefore a right angle and hence, by the
definition of perpendicular, 𝑷𝑹 ⊥ 𝑴𝑸 . Since rays 𝑷𝑹 ⊥ 𝑷𝑸 are
contained in 𝑷𝑹 𝒂𝒏𝒅 𝑴𝑸 , respectively, therefore 𝑷𝑹 ⊥ 𝑴𝑸 .
13. Part 2: Show that 𝑷𝑹 𝒊𝒔 𝒕𝒉𝒆 𝒐𝒏𝒍𝒚 𝒍𝒊𝒏𝒆 ⊥ 𝑴𝑸 through P.
There is only one line 𝑷𝑹 ⊥ 𝑴𝑸 or other than 𝑷𝑹 there is another line 𝑷𝑹 ⊥
𝑴𝑸 . Assume that there is another line 𝑷𝑺 ⊥ 𝑴𝑸 . Then by the definition of
perpendicular, ∠𝑺𝑷𝑸 is a right angle and therefore 𝒎∠𝑹𝑷𝑸 = 𝟗𝟎. This
contradicts the Angle Construction Postulate.
Since the assumption leads to contradiction of postulate, it must be false.
Therefore, there is only one line 𝑷𝑹 ⊥ 𝑴𝑸 .
R
S
M P Q
Given: MQ and point P on MQ
Prove: a. There is at least on line PR ⊥ MQ
b. There is only one line PR ⊥ MQ
14. 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟒𝟏
There is one and only one line
perpendicular to a given line
through an external point.
For Example: 𝐼𝑛 𝑡ℎ𝑒 𝑓𝑖𝑔𝑢𝑟𝑒, 𝑝𝑜𝑖𝑛𝑡 𝑀 𝑙𝑖𝑒𝑠 𝑜𝑛 𝐻𝐷, 𝑡ℎ𝑒 𝑏𝑖𝑠𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 ∠𝐾𝐻𝑁.
If 𝐾𝐻 = 4𝑥 − 15 and 𝑁𝐻 = 2𝑥 + 3. Find 𝑁𝐻.
SOLUTION:
𝑲𝑯 = 𝑵𝑯
4𝑥 − 15 = 2𝑥 + 3 Substitute the value of 𝐾𝐻 and 𝑁𝐻.
4𝑥 − 2𝑥 = 3 + 15 Combine similar terms.
2𝑥 = 18 Simplify.
2𝑥
2
=
18
2
Divide both side by 2.
𝑥 = 9
𝑁𝐻 = 2𝑥 + 3
𝑁𝐻 = 2 9 + 3 Substitute the value of x
𝑁𝐻 = 18 + 3 Simplify.
𝑁𝐻 = 21
H
D
N
M
K