Question

In the figure, $\angle BCD=\angle ADC$ and $\angle ACB=\angle BDA$. Prove that AD = BC and $\angle A=\angle B$.

Answer 1

We have $\angle 1=\angle 2$ and $\angle 3=\angle 4$

$\Rightarrow \angle 1+\angle 3=\angle 2+\angle 4$

$\Rightarrow \angle ACD=\angle BDC$.
Thus in triangles ACD and BDC, we have,

$\angle ADC=\angle BCD$ (given);
CD = CD (common);

$\angle ACD=\angle BDC$ (proved).

By ASA condition $△ACD\cong △BDC$. Therefore

$AD=BC$ and $\angle A=\angle B$.