Question

$D$ is a point on side $BC$ of $△ABC$ such that $AD=AC$.Show that $AB>AD$

Answer 1

Given:$AD=AC$

Angles opposite to equal sides are equal

So, $\angle ACD=\angle ADC$ .......$\left(1\right)$

Now, $\angle ADC$ is an exterior angle of $△ABD$

$\angle ADC=\angle ABD+\angle BAD$ since exterior angle is sum of interior opposite angles.

So, $\angle ADC>\angle ABD$

$\Rightarrow \angle ACD>\angle ABD$ since side opposite to the greater angle is longer

$\therefore AB>AD$ as $AC=AD$ given

Hence proved.