# Problem 8. (5 points)

In triangle $ABC$, $AB=13$ and $BC=15$. A point $D$ is taken on side $AC$ such that $AD=5$ and $CD=9$. The bisector of the angle adjacent to angle $A$ intersects line $BD$ at point $E$. Find $DE$.
Answer: $7.5 \| 15 / 2$

## Solution:

Let's calculate the area of triangle $ABC$ using Heron's formula: $S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{21 \cdot 6 \cdot 7 \cdot 8}=84$ (where $p$ is the semi-perimeter of the triangle, and $a, b, c$ are its sides). Then the length of the height $BH$, dropped to side $AC$, is $\frac{2S}{AC}=12$, from which, by the Pythagorean theorem, $AH=\sqrt{AB^2-BH^2}=5=AD$. Thus, $D$ and $H$ are the same point, and $BD$ is the height.

Let $AK$ be the angle bisector of angle $K$ in triangle $ABD$. Then, by the main property of the angle bisector, $BK:KD=AB:AD=13:5$, from which $KD=\frac{5}{18} \cdot BD=\frac{10}{3}$.

The angle $KAE$, formed by the angle bisectors of adjacent angles, is a right angle, meaning $AD$ is the height in the right triangle $KAE$ and $AD^2=KD \cdot DE$, from which $DE=25:\frac{10}{3}=7.5$.