Question

In the given figure $AD=CD$ and $BD=AD$.

The measure of $\angle ADC$ in degrees is

Answer 1

$\angle ACE+\angle ACD={180}^o$
$130+\angle ACD={180}^o$
$\angle ACD={50}^o$
Since, $AD=DC$,
thus, $\angle DAC=\angle ACD={50}^o$
In $△ADC$,
$\angle DAC+\angle ACD+\angle ADC={180}^o$
$50+50+\angle ADC={180}^o$
$\therefore \angle ADC={80}^{\circ }$