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Question

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

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Solution

Given : An isosceles ABC having AB = AC. AP is the bisector of ext CAD and CP || AB.
To Prove : PAC=BCA and ABCP is a parallelogram
Proof : In ABC, we have
AB = AC [Given]
1=2 ....(1)
[Angles opposite to equal sides in a are equal]
Now, in ABC, we have
ext CAD=1+2
[Exterior angle of a trinagle is equal to the sum of two opposite interior opposite angles]


ext CAD=22 [1=2(from(1))] 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 parallel. Hence, ABCP is a parallelogram.

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