In figure ABC is an isosceles triangle in which AB = AC. CP || AB and AP is the bisector of exterior ∠CADof△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=∠BCAandABCP 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=2∠2[∠1=∠2(from(1))]⇒2∠3=2∠2
[AP is the bisector of ext. ∠CAD∴∠CAD=2∠3]⇒∠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.