In a right triangle $A B C$ with a right angle at $B$, the bisector of angle $A$ intersects side $B C$ at point $D$. It is known that $B D=4, D C=6$. Find the area of triangle $A D C$.

#
Apply the property of the bisector of a triangle

## Solution

By the property of the bisector of a triangle

$$
\frac{A B}{A C}=\frac{B D}{D C}=\frac{2}{3}
$$

Let $A B=2 x, A C=3 x$. By the Pythagorean theorem

$$
B C^{2}+A B^{2}=A C^{2} \text {, or } 100+4 x^{2}=9 x^{2} .
$$

From this, we find that $x=2 \sqrt{5}$. Then

$$
A B=4 \sqrt{5}, S_{\triangle \mathrm{ABC}}=\frac{1}{2} B C \cdot A B=20 \sqrt{5}
$$

Therefore,

$$
S_{\triangle \mathrm{ADC}}=\frac{D C}{B C} S_{\Delta \mathrm{ABC}}=\frac{3}{5} S_{\Delta \mathrm{ABC}}=12 \sqrt{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-11.jpg?height=323&width=644&top_left_y=1371&top_left_x=707)

Answer

$12 \sqrt{5}$.