In a △ ABC , D and E are points on the sides AB and AC respectively such that BD = CE. If ∠ B = ∠ C , then show that…

In $\triangle ABC$, $$\begin{aligned}& \angle B = \angle C \\ & \Rightarrow AC = AB \dots (1) \\ & Given, BD = CE \dots (2) \\ & Subtract (2) from (1), we have \\ & AD = AE \dots (3) \\ & (3) \div (2)\end{aligned}$$, we have $$\begin{aligned}& \frac{AD}{BD} = \frac{AE}{CE} \\ & Therefore, DE \parallel BC\end{aligned}$$.