Task A-3.6. (10 points)

In a right-angled triangle, the bisector of one acute angle divides the opposite leg into segments of length 4 and 5. What is the area of this triangle?
## First Solution.

Let $ABC$ be a right triangle with a right angle at vertex $C$, and let the angle bisector of $\angle BAC$ intersect side $\overline{BC}$ at point $D$. Denote $\angle BAC = \alpha$ and $|AC| = b$. It is given that $|BC| = 9$.

One of the following possibilities holds:

$$
|DC| = 4, |BD| = 5 \quad \text{or} \quad |DC| = 5, |BD| = 4
$$

Let $|DC| = w$ (where $w = 4$ or $w = 5$).

From the right triangle $ABC$, we have

$$
\tan \alpha = \frac{|BC|}{|AC|} = \frac{9}{b}
$$

From the right triangle $ACD$,

$$
\tan \frac{\alpha}{2} = \frac{|DC|}{|AC|} = \frac{w}{b}
$$

Since

$$
\tan \alpha = \tan\left(2 \cdot \frac{\alpha}{2}\right) = \frac{2 \tan \frac{\alpha}{2}}{1 - \left(\tan \frac{\alpha}{2}\right)^2}
$$

It follows that

$$
\frac{9}{b} = \frac{2 \cdot \frac{w}{b}}{1 - \left(\frac{w}{b}\right)^2}
$$

$$
\frac{9}{b} \left(1 - \frac{w^4}{b^2}\right) = \frac{2w}{b}, \quad 1 - \frac{w^2}{b^2} = \frac{2w}{9}, \quad b^2 = \frac{w^2}{1 - \frac{2w}{9}}
$$

For $w = 5$, the equation has no solution,

and for $w = 4$, we get $b = 12$.

Now, $P(ABC) = \frac{1}{2} \cdot |AC| \cdot |BC| = \frac{1}{2} \cdot 9 \cdot 12 = 54$.

Note. A student who solves the problem in this way and does not show (and justify) that $|BD| > |CD|$ loses 2 points.

![](https://cdn.mathpix.com/cropped/2024_05_30_d6d90a155ebaa296cbbeg-20.jpg?height=560&width=731&top_left_y=2133&top_left_x=660)