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Question
The diagonal AC of a parallelogram ABCD intersects DP at the point Q, where P is any point on side AB. Prove that CQ x PQ = QA x QD.
Solution
In ΔPQA and ΔDQC
∠PQA = ∠DQC ...(vertically opposite angles)
∠APQ = ∠QDC ...(alternate angles since AB || DC)
Therefore, ΔPQA ∼ ΔDQC
∴ `"CQ"/"QD" = "QA"/"PQ"`
⇒ CQ x PQ = QA x QD.
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