Given,
In trapezium ABCD, AB || DC, DC produced to Q and L is the midpoint of BC.
∴ BL = CL
Proof:
DC produced to Q and AB || DC.
So, DQ || AB.
In ΔCLQ and ΔBLP,
CL = BL [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) …(i)
Now, ar(ABCD)=ar(APQD)−ar(ΔCQL)+ar(ΔBLP)
=ar(APQD)−ar(ΔBLP)+ar(ΔBLP) [from Eq. (i)]
⇒ ar (ABCD) = ar (APQD)