Given: ABCD is a rhombus, DPR and CBR are straight lines. Prove that : DP $\times$ CR $=$ DC $\times$ PR

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Solution

$A B C D$ is a rhombus. $D P R$ and $C B R$ are straight lines.
$∴$ $A D ∥ C R$
In $△ A P D$ and $△ C P R$
$\Rightarrow$ $∠ A P D = ∠ C P R$ [ Vertically opposite angles ]
$\Rightarrow$ $∠ D A P = ∠ P C R$ [ Alternate angles ]
$∴$ $△ A P D \sim △ C P R$ [ By AA criterion ]
Since, corresponding sides of similar triangles are proportional.
$\frac{D P}{P R} = \frac{A D}{C R} = \frac{A P}{C P}$

$\Rightarrow$ $\frac{D P}{P R} = \frac{A D}{C R}$

$\Rightarrow$ $\frac{D P}{P R} = \frac{D C}{C R}$ [ Since, $D C = A D,$ as $A B C D$ is a rhombus ]
$\Rightarrow$ $D P \times C R = D C \times P R$

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Similar questions

Given : ABCD is a rhombus, DPR and CBR are straight lines. Prove that : DP $\times$ CR = DC $\times$ PR.

ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.

A B C D

\frac{D P}{P Q} = \frac{D C}{B Q}

A B C D

\frac{D Q}{D P} = \frac{A Q}{D C}

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