In fig. ABC is an isosceles triangle in which AB=AC.CP||ABandAP 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 ∠CADandCP||AB.
To Prove : ∠PAC=∠BCAandABCP
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] ⇒ext∠CAD=2∠2[∠1=∠2(from(i))] ⇒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 equal. Hence, ABCP is a parallelogram.