Question

ABC is an isosceles triangle in which $AB=AC$. $AD$ is the bisector of exterior angle $PAC$ and $CD$ is parallel to $AB$. Prove that
(i) $\angle DAC=\angle BCA$
(ii) $ABCD$ is parallelogram.

Answer 1

AD bisects $\angle PAC$, then

$\angle PAD=\angle DAC=\frac{1}{2}\angle PAC..............\left(1\right)$

Also $AB=AC$

$\therefore \angle BCA=\angle ABC...........\left(2\right)$

In $△$ ABC, $\angle PAC$ is an exterior angle.

$\angle PAC=\angle ABC+\angle BCA$

$\angle PAC=\angle BCA+\angle BCA$ [from (2)]

$\angle PAC=2\angle BCA$

$\angle BCA=\frac{1}{2}\angle PAC$

$\angle BCA=\angle DAC$ [from (1)]

$\left(ii\right)$ For lines $BC$ and $AD$ , $AC$ is transversal & $\angle DAC$ &

$\angle BCA$ are alternate interior angles and are equal. Therefore $BC\parallel AD$. In ABCD, $BC\parallel AD$ & $AB\parallel CD$ .

Since, opposite sides are parallel. ABCD is a parallelogram.