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 $\angle DAC=\angle BCA$

Answer 1

$△ABC$ where $AB=AC,AD$ bisect $\angle QAC$ and $CD||AB$

Hence, $\angle PAD=\angle DAC=\frac{1}{2}\angle PAC$ .....(i)

Also given,

$AB=AC$

$\therefore \angle BCA=\angle ABC$ [Angles opposite to equal sides are equal]

For $△ABC$,

$\angle PAC$ is an exterior angle.

$\therefore \angle PAC=\angle ABC+\angle BCA$ ...... [Exterior angle is sum of interior opposite angles]

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

$\Rightarrow \angle PAC=2\angle BCA$

$\therefore \frac{1}{2}\angle PAC=\angle BCA$ ....... [Since $\angle ABC=\angle BCA$]

$\therefore \angle BCA=\frac{1}{2}\angle PAC$ ....... [Since $\angle DAC=\frac{1}{2}\angle PAC$]

$\Rightarrow \angle BCA=\angle DAC$