In a trapezium ABCD, AB || DC and M is the midpoint of BC. Through M,

1.	In a trapezium ABCD, AB \|\| DC and M is the midpoint of BC. Through M, a line PQ \|\| AD has been drawn which meets AB in P and DC produced in Q, as shown in the adjoining figure. Prove that ar (ABCD) = ar (APQD).
Answer» Consider △ MCQ and △ MPB From the figure we know that ∠ QCM and ∠ PBM are alternate angles So we get ∠ QCM = ∠ PBM We know that M is the midpoint of BC CM = BM ∠ CMQ and ∠ PBM are vertically opposite angles ∠ CMQ = ∠ PBM By ASA congruence criterion △ MCQ ≅ △ MPB So we get Area of △ MCQ = Area of △ MPB We know that Area of ABCD = Area of APQD + Area of DMPB – Area of △ MCQ So we get Area of ABCD = Area of APQD Therefore, it is proved that ar (ABCD) = ar (APQD).

In a trapezium ABCD, AB || DC and M is the midpoint of BC. Through M, a line PQ || AD has been drawn which meets AB in P and DC produced in Q, as shown in the adjoining figure. Prove that ar (ABCD) = ar (APQD).

Answer»

Consider △ MCQ and △ MPB

From the figure we know that

∠ QCM and ∠ PBM are alternate angles

So we get

∠ QCM = ∠ PBM

We know that M is the midpoint of BC

CM = BM

∠ CMQ and ∠ PBM are vertically opposite angles

∠ CMQ = ∠ PBM

By ASA congruence criterion

△ MCQ ≅ △ MPB

So we get

Area of △ MCQ = Area of △ MPB

We know that

Area of ABCD = Area of APQD + Area of DMPB – Area of △ MCQ

So we get

Area of ABCD = Area of APQD

Therefore, it is proved that ar (ABCD) = ar (APQD).