$P$ is the mid-point of the side $CD$ of a parallelogram $ABCD$. $A$ line through $C$ parallel to $PA$ intersects $AB$ at $Q$ and $DA$ produced at $R$. Prove that $DA = AR$ and $CQ = QR$.

(N/A) $ABCD$ is a parallelogram. $P$ is the mid-point of $CD$. $A$ line through $C$ parallel to $PA$ intersects $AB$ at $Q$ and $DA$ produced at $R$.
In $\Delta DCR$,$P$ is the mid-point of $CD$ and $AP \parallel CR$ (since $AP \parallel CQ$ and $Q$ lies on $CR$).
Therefore,by the converse of the Mid-point Theorem,$A$ is the mid-point of $DR$,which implies $DA = AR$.
Now,consider $\Delta ARQ$ and $\Delta BCQ$:
$1$. $AR = BC$ (Since $AD = AR$ as proved above,and $AD = BC$ as opposite sides of a parallelogram).
$2$. $\angle 1 = \angle 2$ (Vertically opposite angles).
$3$. $\angle 3 = \angle 4$ (Alternate interior angles,as $AB \parallel DC$ and $RC$ is a transversal).
Therefore,$\Delta ARQ \cong \Delta BCQ$ by the $ASA$ congruence rule.
Thus,$CQ = QR$ by $CPCT$ (Corresponding Parts of Congruent Triangles).
Hence,$DA = AR$ and $CQ = QR$ is proved.