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Given, $ABCD$ is a parallelogram.

$AC$ is a diagonal intersects $DP$ at $Q$ and $P$ is a point on side $AB$.

Now, we have drawn a figure, according to the given situation in question.

To prove: $CQ \times PQ = QA \times QD$.

In $\vartriangle CQD$and $\vartriangle APQ$, we have;

$\angle CQD = \angle AQP$ (Vertically opposite angles and AB || CD)

$\angle QCD = \angle QAP$ (Alternate angles)

Thus, $\vartriangle CQD \sim \vartriangle APQ$(both the triangle is similar by angle-angle criterion)

$ \Rightarrow \dfrac{{PQ}}{{QD}} = \dfrac{{QA}}{{CQ}}$

[If two triangles are similar, then the ratio of their corresponding sides are equal]

$ \Rightarrow PQ \times CQ = QA \times QD$