Question

In the given figure $PQRS$ is a rhombus such that $RQ$ is produced to $E$ and $SE$ is joined. $SE$ and $PR$ intersect at $D.$ Prove that $SD.RE=SR.DE$

Answer 1

In $△PDS$ and $△RDE,$

$\angle PDS=\angle RDE$ [Vertically opposite angles]

$\angle PSD=\angle RED$ [Corresponding interior angles as $PS\parallel RS$ and $SE$ is the transversal]

$\angle SPD=\angle DRE$ [Corresponding interior angles]

So $\Delta PDS\sim \Delta RDE$ [By $AAA$ criteria for similarity]

So $\frac{SD}{DE}=\frac{PS}{RE}=\frac{SR}{RE}$ [All the sides of rhombus $PQRS$ are equal]

So, $SD\times RE=SR\times DE$