Question

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Figure). Prove that ar (ABCD) = ar (APQD)

Answer 1

Given: In trapezium ABCD, AB || DC, DC produced in Q and L is the mid-point of BC.

∴ BL = CL

To prove: ar (ABCD) = ar (APQD)

Proof: Since, DC produced in Q and AB || DC.

So, DQ || AB

In ΔCLQ and ΔBLP,

CL = BL   ...[Since, L is the mid-point of BC]

∠LCQ = ∠LBP   ...[Alternate interior angles as BC is a transversal]

∠CQL = ∠LPB  ...[Alternate interior angles as PQ is a transversal]

∴ ΔCLQ ≅ ΔBLP   ...[By AAS congruence rule]

Then, ar (ΔCLQ) = ar (ΔBLP)  [Since, congruent triangles have equal area]  ...(i)

Now, ar (ABCD) =  ar (APQD) – ar (ΔCQL) + ar (ΔBLP)

= ar (APQD) – ar (ΔBLP) + ar (ΔBLP)  ...[From equation (i)]

⇒ ar (ABCD) = ar (APQD)

Hence proved.