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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.

=> 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