Question

In the given triangle $ABC,$ find the measure of an $\angle ADC$.

Answer 1

Given $\angle ACE={130}^o$
$\angle ACE+\angle ACD={180}^o$
$\angle ACD={180}^o-{130}^o={50}^o$
Given , $AD=CD$

Hence, $\angle CAD=\angle ACD=50$ ......(Isosceles triangle Property)
$\angle ADC+\angle CAD+\angle ACD={180}^o$ ........(Angle sum property)
$\angle ADC+{50}^o+{50}^o={180}^o$
$\therefore \angle ADC={80}^{\circ }$