Question

In figure, $AD=BD$ and $\angle C=\angle E$. Prove that $BC=AE$.

Answer 1

In $\Delta AED$ and $\Delta BCD$
$AD=BD$ (given)
$\angle ADE=\angle BDC$ (vertically opposite angles)
$\angle E=\angle C$
$\therefore \Delta AED\cong \Delta BCD$ (by AAS congruency property)
Similarly, $AE=BC$ (by c.p.c.t)