Question

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

Answer 1

In $\Delta DAC$,

$AD=AC$ (Given)

So, $\angle ADC=\angle ACD$ (Angles opposite to equal sides)

Now, $\angle ADC$ is an exterior angle for $△ABD$.

So, $\angle ADC>\angle ABD$

$\angle ACD>\angle ABD$

$\angle ACB>\angle ABC$

$AB>AC$ (Side opposite to larger angle in $△ABC$)

$AB>AD$ $(AD=AC)$ $\left[henceproved\right]$