Problem 2. The height drawn to the hypotenuse of the right triangle $A B C,\left(\measuredangle C=90^{\circ}\right)$ divides the hypotenuse into two segments with lengths $\overline{A C_{1}}=9 \mathrm{~cm}$ and $\overline{C_{1} B}=16 \mathrm{~cm}$. A line is drawn from vertex $A$ of the triangle passing through the midpoint of the height $C C_{1}$. Determine the length of that part of the line that lies within the triangle.
Solution. The height is the geometric mean of the segments of the hypotenuse, i.e. $h^{2}=9 \cdot 16 ; \quad h=12 \mathrm{~cm}$. From $\triangle A C_{1} P$ we have

$$
\begin{aligned}
& \overline{A P}^{2}=\overline{A C}^{2}+{\overline{C_{1} P}}^{2} \\
& \overline{A P}^{2}=9^{2}+\left(\frac{1}{2} \cdot 12\right)^{2} ; \overline{A P}=3 \sqrt{13}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_1d413d45628348221ff6g-6.jpg?height=320&width=538&top_left_y=1765&top_left_x=940)

Let $P Q$ be the midline of $\triangle C C_{1} B$.

Since $\triangle A B E \sim \triangle P Q E$ (they have equal angles), we get

$$
\overline{A E}: \overline{P E}=\overline{A B}: \overline{P Q} ;(3 \sqrt{3}+\overline{P E}): \overline{P E}=25: 8 ; \overline{P E}=\frac{24 \sqrt{13}}{17} ; \overline{A E}=\frac{75 \sqrt{13}}{17}
$$