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# Let P-Q and R be the three vertices of a triangle- Prove that the line.docx

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# Let P-Q and R be the three vertices of a triangle- Prove that the line.docx

Let P,Q and R be the three vertices of a triangle. Prove that the line
segment that joints the midpoints of two of the sides is parallel to and half the length
of the third side.
Solution
in your triangle ABC, R is the mid point of AB and S is the mid point of AC. Now the basic proportionality theorem recalls the fact that when a line segment is drawn parallel to the base of the triangle say RS drawn parallel to BC, then AR/RB = AS/AC Using the converse of basic proportionality theorem, we get AR/AB = 1 as R is the mid point of AB. And AS/AC is also equal to 1 as S is the mid point of AC. Hence AR/RB = AS/AC establishes by converse of BPT theorem, RS becomes parallel to AB. So first part of the problem has been proved. ie the line segment drawn is || to the third side. For second part, as AR/AB = 1/1 by ratio and proportionality we get AR /(AR+RB = 1/2 ====> AR /AB = 1/2 Same way, we can prove AS/AC = 1/2 NOw in the two triangles ABC and ARS, by SAS rule the two triangles become similar. Hence RS / AB = 1/2 Or RS = AB/2 So second part is also proved. ie the segment is half of the third side.
.

Let P,Q and R be the three vertices of a triangle. Prove that the line
segment that joints the midpoints of two of the sides is parallel to and half the length
of the third side.
Solution
in your triangle ABC, R is the mid point of AB and S is the mid point of AC. Now the basic proportionality theorem recalls the fact that when a line segment is drawn parallel to the base of the triangle say RS drawn parallel to BC, then AR/RB = AS/AC Using the converse of basic proportionality theorem, we get AR/AB = 1 as R is the mid point of AB. And AS/AC is also equal to 1 as S is the mid point of AC. Hence AR/RB = AS/AC establishes by converse of BPT theorem, RS becomes parallel to AB. So first part of the problem has been proved. ie the line segment drawn is || to the third side. For second part, as AR/AB = 1/1 by ratio and proportionality we get AR /(AR+RB = 1/2 ====> AR /AB = 1/2 Same way, we can prove AS/AC = 1/2 NOw in the two triangles ABC and ARS, by SAS rule the two triangles become similar. Hence RS / AB = 1/2 Or RS = AB/2 So second part is also proved. ie the segment is half of the third side.
.

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### Let P-Q and R be the three vertices of a triangle- Prove that the line.docx

1. 1. Let P,Q and R be the three vertices of a triangle. Prove that the line segment that joints the midpoints of two of the sides is parallel to and half the length of the third side. Solution in your triangle ABC, R is the mid point of AB and S is the mid point of AC. Now the basic proportionality theorem recalls the fact that when a line segment is drawn parallel to the base of the triangle say RS drawn parallel to BC, then AR/RB = AS/AC Using the converse of basic proportionality theorem, we get AR/AB = 1 as R is the mid point of AB. And AS/AC is also equal to 1 as S is the mid point of AC. Hence AR/RB = AS/AC establishes by converse of BPT theorem, RS becomes parallel to AB. So first part of the problem has been proved. ie the line segment drawn is || to the third side. For second part, as AR/AB = 1/1 by ratio and proportionality we get AR /(AR+RB = 1/2 ====> AR /AB = 1/2 Same way, we can prove AS/AC = 1/2 NOw in the two triangles ABC and ARS, by SAS rule the two triangles become similar. Hence RS / AB = 1/2 Or RS = AB/2 So second part is also proved. ie the segment is half of the third side.