Question

$ABC$ is an isosceles triangle in which $AB=AC.AD$ bisects exterior angle $QAC$ and $CD\parallel BA$ as shown in the figure. Show that:
(i) $\angle DAC=\angle BCA$
(ii) $ABCD$ is a parallelogram

Answer 1

Given : $ABC$ is an isosceles triagle in which $AB=AC.AD$ bisects exterior angle $QAC$ and $CD||BA$.

To show :

(i)$\angle DAC=\angle BCA$

(ii) $ABCDABCD$ is a parallelogram

Proof :

(i)

$\angle ABC=\angle BCA=y\left(let\right)$ (Because triangle $ABC$ is an isosceles triangle)

$\angle QAD=\angle DAC=x\left(let\right)$ (Given)

$\angle DCA=\angle BAC=z\left(let\right)$ (Alternate interior angles)

And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.

So,

$\angle QAD+\angle DAC=\angle ABC+\angle BCA$

$x+x=y+y$

$2x=2y$

$x=y$

$\angle DAC=\angle BCA$ (hence proved)

(ii)

Now because,

$\angle DAC=\angle BCA$ (proved above)

Therefore , $AD||BC$

And $CD||BA$ (Given)

Since opposite sides of quadrilateral $ABCD$ are parallel therefore $ABCD$ is a parallelogram.