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 × PQ = QA × QD.

Answer 1

Given: □ ABCD is a parallelogram. 

P is a point on AB. 

DP and AC intersect at Q. 

R.T.P.: CQ . PQ = QA . QD. 

Proof: In △CQD, △AQP 

∠QCD = ∠QAP 

∠CQD = ∠AQP 

∴ ∠QDC = ∠QPA 

(∵ Angle sum property of triangles) 

Thus, △CQD ~ △AQP by AAA similarity condition.

\(\frac{CQ}{AQ}=\frac{QD}{QP}=\frac{CD}{AP}\)

[∵ Ratio of corresponding sides of similar triangles are equal]

\(\frac{CQ}{AQ}=\frac{QD}{QP}\)

CQ . PQ = QA . QD [Q.E.D]