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 :

The diagonal AC of a parallelogram

ABCD intersects DP at the point Q ,

where ' P ' is any point on side AB .

To Prove :

CQ × PQ = QA × QD

proof :

In ∆APQ and ∆CDQ

<AQP = <CQD

( vertically opposite angles )

<QAP = <QCD

[ alternate angles ]

Therefore ,

∆APQ ~ ∆CQD

[ A.A criterion ]

=> PQ/QD = QA/CQ

CQ × PQ = QA × QD

Hence , proved .

I hope this helps you.

: )