Question

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 1

In △MQC and △MPB,
MC = MB
(M is the midpoint of BC)
∠CMQ = ∠BMP
(Vertically opposite angles)
∠MCQ = ∠MBP
(Alternate interior angles on the parallel lines AB and DQ)
Thus, △MQC ≅ △MPB (ASA congruency)
⇒ar(△MQC) = ar(△MPB)
⇒ar(△MQC) + ar(APMCD) = ar(△MPB) + ar(APMCD)
⇒ar(APQD) = ar(ABCD)