In the figure,$PSDA$ is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ = QR = RS$ and $PA \parallel QB \parallel RC$. Prove that $\operatorname{ar}(PQE) = \operatorname{ar}(CFD)$.

(A) $PSDA$ is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ = QR = RS$ and $PA \parallel QB \parallel RC$.
We have to prove that $\operatorname{ar}(\triangle PQE) = \operatorname{ar}(\triangle CFD)$.
Since $PSDA$ is a parallelogram,$PS = AD$ and $PS \parallel AD$.
Given $PQ = QR = RS$,we have $PQ = QR = RS = \frac{1}{3} PS$.
Since $PA \parallel QB \parallel RC \parallel SD$,the segments $AB, BC, CD$ are also equal by the Intercept Theorem.
Thus,$AB = BC = CD = \frac{1}{3} AD$.
Therefore,$PQ = CD$.
Now,consider $\triangle PQE$ and $\triangle CFD$:
$1$. $PQ = CD$ (Proved above).
$2$. $\angle QPE = \angle FDC$ (Alternate interior angles,as $PS \parallel AD$ and $PD$ is a transversal).
$3$. $\angle PQE = \angle FCD$ (Corresponding angles,as $QB \parallel RC$ and $PS$ is a transversal).
By the $ASA$ congruence rule,$\triangle PQE \cong \triangle CFD$.
Since congruent triangles have equal areas,$\operatorname{ar}(\triangle PQE) = \operatorname{ar}(\triangle CFD)$.