Question

In Figure $ PSDA$ is a parallelogram.

Points $ Q$ and $ R$ are taken on $ PS$ such that $ PQ = QR = RS$ and $ PA \left|\right| QB \left|\right| RC$. Prove that $ ar \left(PQE\right) = ar \left(CFD\right)$.

Answer 1

Prove $ar\left(PQE\right)=ar\left(CFD\right)$

According to the given data

$PA║QB║RC║SD,PQ=QR=RS$

From the Equal intercept theorem, if a transversal makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts.

So it can be written like

$PE=EF=FD$ and $AB=BC=CD$

From $△PQE$ &$△DCF$,

$$\begin{gathered} \angle PEQ=\angle DFC\left(\because alternateexteriorangles\right) \\ PE=DF \\ \angle QPE=\angle CDF\left(\because alternateinteriorangles\right) \end{gathered}$$

Hence $△PQE\cong △DCF$

As congruent figures are equal in area

$\Rightarrow ar(△PQE)=ar(△DCF)$

Hence it is proved that $ar(△PQE)=ar(△DCF)$