$ABCD$ is a parallelogram and $P$ and $Q$ are points on the diagonal $AC$ such that $AP = PQ = QC$. Prove that $BQ \parallel DP$ and $BD$ bisects $PQ$.

(N/A) $1$. In $\triangle ABP$ and $\triangle CDQ$:
$AB = CD$ (Opposite sides of parallelogram $ABCD$ are equal).
$\angle BAP = \angle DCQ$ (Alternate interior angles as $AB \parallel CD$ and $AC$ is a transversal).
$AP = CQ$ (Given).
By $SAS$ congruence criterion, $\triangle ABP \cong \triangle CDQ$.
Thus, $BP = DQ$ and $\angle ABP = \angle CDQ$.
$2$. In $\triangle ADP$ and $\triangle CBQ$:
$AD = CB$ (Opposite sides of parallelogram $ABCD$ are equal).
$\angle DAP = \angle BCQ$ (Alternate interior angles as $AD \parallel BC$ and $AC$ is a transversal).
$AP = CQ$ (Given).
By $SAS$ congruence criterion, $\triangle ADP \cong \triangle CBQ$.
Thus, $DP = BQ$ and $\angle ADP = \angle CBQ$.
$3$. Since $BP = DQ$ and $DP = BQ$, quadrilateral $PBQD$ is a parallelogram (opposite sides are equal).
Therefore, $BQ \parallel DP$.
$4$. In parallelogram $PBQD$, diagonals $BD$ and $PQ$ bisect each other. Hence, $BD$ bisects $PQ$.