Question

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

Answer 1

Given : An isosceles ΔABC having AB = AC.AP is the bisector of ext $\angle CADandCP||AB$.
To Prove : $\angle PAC=\angle BCAandABCP$
Proof : In $\Delta ABC$, we have
AB = AC [Given]
$\Rightarrow \angle 1=\angle 2....\left(i\right)$
[ Angles opposite to equal sides in a $\Delta$ are equal]
Now, in $\Delta ABC$, we have
ext $\angle CAD=\angle 1+\angle 2$
[ Exterior angle of a triangle is equal to the sum of two opposite interior opposite angles]
$\Rightarrow ext\angle CAD=2\angle 2[\angle 1=\angle 2(from\left(i\right)\left)\right]$
$\Rightarrow 2\angle 3=2\angle 2$
[AP is the bisector of ext. $\angle CAD\therefore \angle CAD=2\angle 3]$
$\Rightarrow \angle 3=\angle 2$
Thus, AC intersects lines AP and BC at A and C respectively such that $\angle 3=\angle 2$ 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.