Question

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

Answer 1

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


$\Rightarrow ext\angle CAD=2\angle 2[\angle 1=\angle 2(from\left(1\right)\left)\right]\Rightarrow 2\angle 3=2\angle 2$
[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 parallel. Hence, ABCP is a parallelogram.