In the given figure, $\displaystyle \Delta \mathrm{ABE} \cong \Delta \mathrm{ACD}$. Prove that $\displaystyle \Delta \mathrm{ADE} \sim \Delta \mathrm{ABC}$.

Given \(\displaystyle \triangle \mathrm{ABE} \cong \triangle \mathrm{ACD}\) \[\therefore \mathrm{AB}=\mathrm{AC} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=1 \] \[\text { and } \mathrm{AE}=\mathrm{AD} \Rightarrow \frac{\mathrm{AD}}{\mathrm{AE}}=1 \] From (i) and (ii), we get \[\frac{A B}{A C}=\frac{A D}{A E} \Rightarrow \frac{A E}{A C}=\frac{A D}{A B} \] & \(\displaystyle \angle \mathrm{A}=\angle \mathrm{A}\) \[\therefore \triangle \mathrm{ADE} \sim \triangle \mathrm{ABC} \]