Question

In trapezium ABCD, AB || DC and L is the midpoint of BD. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. Prove that ar(trap. ABCD) = ar(||gm APQD).

Answer 1

Given: A trapezium ABCD, in which AB ||​ DC and L is the mid point of BC (i.e., BL = CL).
Now, in ∆LPB and ​∆LQC, we have:
BL = CL (Given)
∠BLP = ∠CLQ (Vertically opposite angles)
∠PBL = ∠QCL (Alternate angles)
∴​ ∆LPB ≅ ​∆LQC (ASA congruency)

Now, ar(trapz.ABCD) = ar (APLCD ) + ar(∆​ LPB)
⇒ ar(trapzABCD) = ar(APLCD ) + ar(∆​ LQC) [∴ (ar (∆​ LPB) = ar(∆​ LQC)]
⇒ ar(trapzABCD) = ar($\parallel \mathrm{gm}$ APQD)
Hence proved.