Problem 4. Given a right triangle $\triangle A B C$, with a right angle at vertex $C$. The angle bisector of the right angle divides the hypotenuse in the ratio 1:2. Determine in what ratio the altitude dropped from vertex $C$ divides the hypotenuse.
Solution. Let

$$
\overline{B C}=a, \overline{A C}=b, \overline{A B}=c
$$

and let $D$ and $E$ be the feet of the angle bisector and the altitude dropped from vertex $C$ to the hypotenuse $A B$, respectively. Then, by the Angle Bisector Theorem, we have

![](https://cdn.mathpix.com/cropped/2024_06_05_a73f0c01d543f9cc2089g-11.jpg?height=262&width=509&top_left_y=508&top_left_x=971)
$\overline{A D}: \overline{D B}=\overline{A C}: \overline{B C}$, from which, due to the given condition, we get

$$
b: a=1: 2
$$

Furthermore, from the Euclidean formulas for $\triangle A B C$, we have $b^{2}=\overline{A E} \cdot c$ and $a^{2}=\overline{B E} \cdot c$, from which we get $\overline{A E}: \overline{B E}=b^{2}: a^{2}$. Now, due to (1), we have $\overline{A E}: \overline{B E}=1: 4$