ABCD is a trapezium with AB parallel to CD. If AC and BD intersect at E and triangle AED is similar to triangle BEC. Prove that AD = BC

We have

Given that, ABCD is a trapezium.

AB ∥ CD.

AC and BD are intersecting at E.

Also, ∆AED ∼ ∆BEC.

To Prove: AD = BC.

Proof: If we prove that ∆AEB ∼ ∆CDE, we can easily prove that AD = BC.

As we have been given that ∆AED ∼ ∆BEC.

So, we can write as

[∵ corresponding sides are proportional in similar triangles] …(i)

Take ∆ABE and ∆CDE.

∠AEB = ∠CED [∵ they are vertically opposite angles]

∠EAB = ∠ECD [∵ they are alternate angles (AB ∥ DC)]

∴ ∆ABE ∼ ∆CDE (by AA Similarity)

And since corresponding sides are equal, we can write as

Take the last two ratios,

…(ii)

From (i) →

From (ii) →

⇒ EC = ED

So, From (i),

⇒ AD = BC