Question

$D$ is a point on side $BC$ of $\Delta ABC$ such that $AD=AC$ (see figure). 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 $\Delta ABD$.
So, $\angle ADC>\angle ABD$
or, $\angle ACD>\angle ABD$
or, $\angle ACB>\angle ABC$
So, $AB>AC$ (Side opposite to larger angle in $\Delta ABC)$
or, $AB>AD(AD=AC)$