## Task B-2.5.

Points $M$ and $N$ are located on sides $\overline{A B}$ and $\overline{A D}$ of parallelogram $A B C D$, respectively. It is given that $\frac{|A M|}{|A B|}=\frac{2}{3}$ and $\frac{|A N|}{|A D|}=\frac{3}{4}$. Segments $\overline{M N}$ and $\overline{A C}$ intersect at point $P$. What is the ratio $\frac{|A P|}{|A C|}$?
## Solution.

Let $a=|A B|, b=|A D|$. It follows that $|A M|=\frac{2}{3} a,|A N|=\frac{3}{4} b$.

![](https://cdn.mathpix.com/cropped/2024_05_30_bbee2d45ecbf3fd03f8eg-09.jpg?height=491&width=1062&top_left_y=1613&top_left_x=423)

Let $S \in \overline{A C}$ such that $N S \| A B$.

Clearly, $\triangle A N S \sim \triangle A D C(\mathrm{AA})$ with a similarity coefficient of $\frac{3}{4}$, so it follows that

$|N S|=\frac{3}{4} a,|A S|=\frac{3}{4}|A C|$.

Since $\varangle A P M=\varangle S P N$ (vertical angles) and $\varangle M A P=\varangle N S P$ (angles with parallel sides), we have $\triangle P S N \sim \triangle P A M$.

Then,

$$
\begin{gathered}
\frac{|A P|}{|S P|}=\frac{|A M|}{|S N|} \\
\frac{|A P|}{|A S|-|A P|}=\frac{\frac{2}{3} a}{\frac{3}{4} a} \\
\frac{|A P|}{\frac{3}{4}|A C|-|A P|}=\frac{8}{9} \\
9|A P|=6|A C|-8|A P| \\
\frac{|A P|}{|A C|}=\frac{6}{17}
\end{gathered}
$$