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

Hint: In this problem, we need to use the property of the similar triangles to prove the given correlation. The corresponding sides of the similar triangles are proportional to each other. All the sides of a rhombus are equal.

Complete step by step answer:
Given: ABCD is a rhombus; DPR and CBR are straight lines. Therefore,
\[AD\parallel BC\]
Now, in \[\Delta APD\] and \[\Delta CPR\],
\[
  \angle APD = \angle CPR\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{vertical opposite angles}}} \right) \\
  \angle DAP = \angle PCR\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Alternate angles}}} \right) \\
\]
Therefore, by angle-angle criterion \[\Delta APD \sim \Delta CPR\].
We know that the corresponding sides of the similar triangles are proportional. Therefore,
\[
  \,\,\,\,\,\,\dfrac{{DP}}{{PR}} = \dfrac{{AD}}{{CR}} = \dfrac{{AP}}{{CP}} \\
   \Rightarrow \dfrac{{DP}}{{PR}} = \dfrac{{AD}}{{CR}} \\
   \Rightarrow \dfrac{{DP}}{{PR}} = \dfrac{{DC}}{{CR}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {DC = AD,ABCD\,\,{\text{is a rhombus}}} \right) \\
   \Rightarrow DP \times CR = DC \times PR \\
\]
Hence, proved.

Note: Two triangles are similar, if their corresponding sides are in proportion and corresponding angles are congruent. A rhombus is said to be a square if any of the angles of the rhombus is the right angle.