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  • P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = QR.

P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = QR.

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Solution.
Given: In a parallelogram ABCD, P is the mid-point of DC
To Prove: DA = AR and CQ = QR

Proof: ABCD is a parallelogram
 BC = AD and BC\parallel AD
Also, DC = AB and DC\parallel AB
P is mid-point of DC
DP=PC=\frac{1}{2}DC
Now QC\parallel AP and PC\parallel AQ
So APCQ is a parallelogram
AQ=PC=\frac{1}{2}DC

\frac{1}{2}AB=BQ         …..(1) {\because DC = AB}
In \triangle AQR  &  \triangle BQC AQ = BQ {from equation 1}
\angle AQR=\angle BQR {vertically opposite angles}
\angle ARQ=\angle BCQ {alternate angles of transversal}
\triangle AQR\cong \triangle BQC {AAS congruence}
AR = BC {by CPCT}
BC = DA
AR = DA
Also, CQ = QR {by CPCT}
Hence Proved .

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