### $ABCD$ is a parallelogram where $P$ and $R$ are the midpoints of sides $BC$ and $DC$ respectively. If the line $PR$ intersects the diagonal $AC$ at $Q$, prove that $AC = 4CQ$.

Step by Step Explanation:
1. Let us draw the image for the situation given in the question.
Also, join $BD$ intersecting $AC$ at $O$. 2. It is given that $P$ is the mid-point of $BC$ and $R$ is the mid-point of $DC$.

Thus, in triangle $CBD$, by using mid-point theorem $PR \parallel BD$.
3. As, \begin{aligned} & PR \parallel BD \\ \implies & PQ \parallel BO \text{ and } QR \parallel OD \end{aligned} Now, in triangle $BCO$, we have $PQ \parallel BO$ and $P$ is the mid point of $BC$.
By the inverse of mid-point theorem, $Q$ is the midpoint of $OC$. $$\implies 2 CQ = OC$$
4. As the diagonals of a parallelogram bisect each other, $AO = OC$.
Thus, \begin{aligned} & AC = AO + OC \\ \implies & AC = 2 OC && [\because \text{ AO = OC]} \\ \implies & AC = 2 \times 2 CQ && [\because \text{ 2 CQ = OC]} \\ \implies & AC = 4 CQ \end{aligned} 