Question

In fig ABCD is a trapezium with AB parallel to DC . If triangle AED is similar to triangle BEC . Prove that AD = BC

Answer 1

hope you get it
it's not to prove that is given
similar by angle angle property

Answer 2

Thank you for asking this question. Here is your answer:

∠1 = ∠2 (these are alternate angles)

∠3 = ∠4 (these are also alternate angles)

and ∠CED = ∠AEB (these are vertically opposite angles)

Δ EDC ≈ ΔEBA

= ED/EB = EC/EA ⇒ ED/EC = EB/EA (equation 1)

We are given this that ΔAED ≈ ΔBEC

ED/EC = EA/EB = AD/BC (this is an equation 2)

So from the equation 1 and equation 2 we will get this:

EB/EA = EA/EB ⇒ (EB)² = (EA)² ⇒ EB = EA

Now we will substitute EB = EA in equation 2, then we will get:

EA/EA = AD/BC ⇒ AD/BC = 1 ⇒ AD = BC

Hence, Proved.

If there is any confusion please leave a comment below.