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.

Answer 1

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.