Question

$PSDA$ is a parallelogram. Points $Q$ and $R$ are takes on $PS$ such that $PQ=QR$ and $PA||QB||RC$. Prove that area $\left(\Delta PQE\right)=\left(\Delta CFD\right)$

Answer 1

Consider $\Delta PQE=\Delta DCF$

Since $AD||PS$ and $AP||BQ||CR||DS$

$\angle E=\angle F$

$EQ=FCPE=DF$

So By $SAS$ criteria

$\Delta PQE$ is congruent to $\Delta CFD$

So $ar\left(\Delta PQE\right)=ar\left(\Delta CFD\right)$