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)
$\angle$CMQ = $\angle$BMP (Vertically opposite angles)
$\angle$MCQ = $\angle$MBP (Alternate interior angles on the parallel lines AB and DQ)
Thus, $△$MQC $\cong$ $△$MPB (ASA congruency)
$\Rightarrow$ar($△$MQC) = ar($△$MPB)
$\Rightarrow$ar($△$MQC) + ar(APMCD) = ar($△$MPB) + ar(APMCD)
$\Rightarrow$ar(APQD) = ar(ABCD)