Question

ABCD is a trapezium, AB parallel to DC. O is the midpoint of BC. Through the point O, PQ parallel to AD has been drawn which intersects AB at Q and DC produced at P. Prove ar(ABCD)=ar(AQPD)

Answer 1

In ΔCOQ and ΔBOP 
∠OCQ = ∠OBP as alternate interior opposite angles

CO = BO as O is the mid point of BC
∠COQ is congruent to ∠BOP (vertically opposite angles)
 so, ΔCOQ is congruent to ΔBOP

so, area of COQ is equal to the area of BOP 
area of ABCOP + area of ΔCOQ = area of ADCOP + area of ΔBOP

Answer 2

Given: ABCD is a trapezium with AB||CD 

L is the midpoint of BC and PQ || AD 

 

Now in ΔCLQ and  ΔBLP 

∠LCQ = ∠LBP 

CL = BL ( L is the midpoint of BC) 

∠CLQ = ∠BLP 

So ΔCLQ = ΔBLP 

Area (ΔCLQ) = Area (ΔBLP) 

Area (ABCLP) + Area (ΔCLQ) = Area (ADCLP) + Area (ΔBLP) 

So area (APQD) = Area (ABCD)