Question

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

Answer 1

ABCD is a trapezium in which AB is parallel to CD. O is mid point of BC.

Through the point O, a line PQ parallel to AD has been dawn which intersects

AB at Q and DC produced at P. Prove that $\operatorname{ar}(\mathrm{ABCD}) = \operatorname{ar}(\mathrm{AQPD})$ .

Solution:

AreaABCD = AreaAQOCD + AreaΔQOB

AreaAQPD = AreaAQOCD + AreaΔPOC

$OC = OB$ (given)

$\angle COP = \angle BOQ$ (vertically opposite angles are equal)

$\angle CPO = \angle BQO$ ( $AB \parallel CD$ - given. If the lines are parallel, then the alternate angles are equal)

So $\Delta QOB = \Delta POC$ and so $AreaABCD = AreaAQPD$ .