Question

In figure, $AB=AC,CD=CA$ and $\angle ADC={20}^0$, find $\angle ABC$.

Answer 1

$CD=CA$
$\angle CAD=\angle ADC={20}^0$
$\angle ACD={180}^0–({20}^0+{20}^0)={140}^0$
$\angle ACB={180}^0–{140}^0={40}^0$
But $AB$ also equal to $AC$
$\angle ABC=\angle ACB={40}^0$