ABCD is a trapezium with AB || DC. AC and BD intersect at E. If △ AED △ BEC , then prove that AD = BC.

Given $$\begin{aligned}& \triangle AED \sim \triangle BEC \\ & \therefore \frac{AE}{BE} = \frac{DE}{CE} = \frac{AD}{BC} \dots (1) \\ & Also AB \parallel DC \Rightarrow \triangle AEB \sim \triangle CED \\ & \therefore \frac{AE}{CE} = \frac{BE}{DE}\end{aligned}$$ or $$\begin{aligned}& \frac{AE}{BE} = \frac{CE}{DE} \dots (2) \\ & From (1)\end{aligned}$$ and $(2)$, we get $$\begin{aligned}& \frac{DE}{CE} = \frac{CE}{DE} \\ & \Rightarrow DE^2 = CE^2 \Rightarrow DE = CE \\ & \therefore\end{aligned}$$ From $(1)$, $\frac{AD}{BC} = 1 \Rightarrow AD = BC$