Question

in the figure in the attachment, given that AB = BE and that BD = BC prove that AE is parallel to DC​

Answer 1

$\frac{BE}{BD}$ = $\frac{BA}{BC}$                                                      {∵ BE = BA and BD = BC (given)}

Consider ΔBEA and ΔBDC

∠EBA = ∠DBC (Vertically opposite Angles)

$\frac{BE}{BD}$ = $\frac{BA}{BC}$  (proved)

∴ ΔBEA ~ ΔBDC

⇒∠BEA = ∠BDC (Corresponding angles in similar triangles are equal)

∠BAE = ∠BCD

But these are also alternate interior angles.

⇒ AE ║ CD

I hope this is correct :)