Q) ABCD is a trapezium with AB ǁ DC. AC and BD intersect at E. If Δ AED ~ Δ BEC, then prove that AD = BC.

Ans:

Let’s start with the diagram for he given question:

Step 1: We are given AB ǁ DC, hence let’s look at △AEB and △CED: 

Here we have: 

∠EDC = ∠EBA (Alternate angles)

∠ECD = ∠BAE (Alternate angles)

and, ∠DEC = ∠AEB (Vertically opposite angles)

∴ △AEB ∼ △CED

⇒

⇒ ….. (i)

Step 2: we are given that △AED ∼ △BEC

∴ ………. (ii)

Step 3: By comparing equations (i) and (ii), we get

⇒ (BE)² = (AE)² 

⇒ BE = AE ……… (iii)

Step 4: From equation (ii), we had:

By substituting BE = AE from equation (iii) in this relation, we get:

1 =

AD = BC

Hence Proved !

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