Question

$D$ is a point in side $BC$ of triangle $ABC$. If $AD>AC$, show that $AB>AD$.

Answer 1

$AD=AC$ ( Given )

So, $\angle ACD=\angle ADC$ [ Angles opposite to equal sides are equal ]

Now $\angle ACD$ is exterior angle of $△ABD$

$\angle ADC=\angle ABD+\angle BAD$ [ exterior angles sum of interior opposite angles ]

So, $\angle ADC>\angle ABD$

$\angle ACD>\angle ABD$ ( from (1))

$AB>AC$ [ side opposite to greater angle is longer ]

$\therefore AB>AD$ ( As $AC=AD$ given )

Hence, proved.