# P is the mi

To Prove: DA = AR and CQ = QR

Given: P is the mid-point of the CD. A line through C parallel to PA intersects AB at Q and DA produced to R.

Concept Used:

Opposite sides of a parallelogram are parallel and equal.

ASA Theorem: If two angles and one side of a triangle is equal to two angles and one side of another triangle, then the triangles are congruent.

Diagram:

Explanation:

Now,

BC = AD [Opposite sides of a parallelogram]

BC || AD [Opposite sides of a parallelogram]

DC = AB and DC || AB [ Opposite sides of a parallelogram]

Now, it is given that P is mid-point of the CD.

DP = PC = 1/2 DC

Now,

QC || AP and PC || AQ,

As the opposite sides are equal and parallel.

APCQ is a parallelogram.

AQ = PC = 1/2 DC = 1/2 AB = BQ

Now, in ΔAQR and BQC,

AQ = BQ

AQR = BQC [Vertically opposite angles]

ARQ = BCQ [Alternate angles]

Therefore, ΔAQR and BQC are congruent by ASA theorem.

AR = BC [By C.P.C.T]

BC = DA

AR = DA

And, CQ = QR

Hence, Proved.

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