Question

In the given figure, if $\angle x=\angle y$ and AB = CB, then prove that AE is equal to CD.

Answer 1

Consider $\Delta ABE$ and $\Delta CBD$.
Given, $\angle x=\angle y$.
$⟹$ ${180}^{\circ }-\angle x={180}^{\circ }-\angle y$
$⟹\angle CDB=\angle AEB$(0.5 Mark)
Now, AB = CB (given) and
$\angle CBD=\angle ABE$ $(\because$ common angles in $\Delta ABE$ and $\Delta CBD$)
Hence, $\Delta ABE\cong \Delta CBD$ (0.5 Mark)
(by AAS criteria of congruency for triangles)
$\therefore$ AE = CD (CPCT) (1 Mark)