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Question
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.
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 || AD
Also, DC = AB and DC || AB
Since, P is the mid-point of DC.
∴ DP = PC = `1/2`DC
Now, QC || AP and PC || AQ
So, APCQ is a parallelogram
∴ AQ = PC = `1/2`DC
= `1/2`AB = BQ [∵ DC = AB] ...(i)
Now, in ΔAQR and ΔBQC,
AQ = BQ ...[From equation (i)]
∠AQR = ∠BQC ...[Vertically opposite angles]
And ∠ARQ = ∠BCQ ....[Alternate interior angles]
∴ ΔAQR = ΔBQC ...[By AAS congruence rule]
∴ AR = BC ...[By CPCT rule]
But BC = DA
∴ AR = DA
Also, CQ = QR ...[By CPCT rule]
Hence proved.
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