Q). In the adjoining figure ABCD is a parallelogram and X is the midpoint of BC. The line AX produced meets BC produced at Q the parallelogram a AQPB is completed. Prove that triangle ABX congruent to triangle QCX and DC is equal to CQ is equal to QP.

In triangle BCP, X is the mid point of line BC and XQ is parallel to PB so Q is the mid point of line CP.
=> CQ = PQ
​Also PQ = AB, so CQ = AB
angle XCQ = angle XBA (AIA)
angle XQC = angle XAB (AIA)
So in triangle QCX and triangle ABX
angle XCQ = angle XBA (AIA)
CQ = AB
angle XQC = angle XAB (AIA)
So by ASA congruency rule triangle ABX is congruent to triangle QCX
​AB = CD and AB = PQ (by 2 //gms)
so CD = PQ