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Question

In fig. ABC is an isosceles triangle in which AB=AC.CP||AB and AP is the bisector of exterior ∠CAD of ΔABC. Prove that PAC=BCA and ABCP is a parallelogram.

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Solution

Given : An isosceles ΔABC having AB = AC.AP is the bisector of ext CAD and CP||AB.
To Prove : PAC=BCA and ABCP
Proof : In ΔABC, we have
AB = AC [Given]
1=2....(i)
[ Angles opposite to equal sides in a Δ are equal]
Now, in ΔABC, we have
ext CAD=1+2
[ Exterior angle of a triangle is equal to the sum of two opposite interior opposite angles]
extCAD=22 [1=2(from (i))]
23=22
[AP is the bisector of ext. CADCAD=23]
3=2
Thus, AC intersects lines AP and BC at A and C respectively such that 3=2 i.e., alternate interior angles are equal.
Therefore,
AP || BC.
Also, CP || AB [Given]
Thus, ABCP is a quadrilateral whose opposite sides are equal. Hence, ABCP is a parallelogram.

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