### In the given figure, points $M$ and $N$ divide the side $AB$ of $\Delta ABC$ into three equal parts. Line segments $MP$ and $NQ$ are both parallel to $BC$ and meet $AC$ at $P$ and $Q$ respectively. Prove that $P$ and $Q$ divide $AC$ into three equal parts i.e $AC = 3AP = 3PQ = 3QC$. A B C X Y M N P Q

Step by Step Explanation:
1. Through $A$, let us draw $XAY || BC$.

Now, $XY || MP || NQ$ are cut by the transversal $AB$ at $A, M,$ and $N$ respectively such that $AM = MN$.

Also, $XY || MP || NQ$ are cut by the transversal $AC$ at $A, P,$ and $Q$ respectively. \begin{aligned} \therefore \space\space AP = PQ &&\ldots\text{(i)} &&[\text{ By intercept theorem }] \end{aligned}
2. Again, $MP || NQ || BC$ are cut by the transversal $AB$ at $M, N,$ and $B$ respectively such that $MN = NB$.

Also, $MP || NQ || BC$ are cut by the transversal $AC$ at $P, Q,$ and $C$ respectively. \begin{aligned} \therefore \space\space PQ = QC &&\ldots\text{(ii)} &&[\text{ By intercept theorem }] \end{aligned} Thus, from $eq \space (\text{i})$ and $eq \space (\text{ii})$, we get $AP = PQ = QC.$

Therefore, $P$ and $Q$ divide $AC$ into three equal parts.

Hence, $AC = 3AP = 3PQ = 3QC$.