3 [Auxiliary area. The area helps to solve the task]

In a right triangle $ABC$, the bisector $CD$ of the right angle $C$ is drawn. It is known that $AD=m, BD=n$. Find the height dropped from vertex $C$.
The bisector of a triangle divides its side into segments proportional to the other two sides, so $\frac{A C}{B C} = \frac{A D}{B D} = \frac{m}{n}$. Let $A C = m x, B C = n x$. By the Pythagorean theorem,

$$
A B^2 = A C^2 + B C^2, (m+n)^2 = m^2 \times 2 + n^2 \times 2
$$

from which we find that $x^2 = \frac{(m+n)^2}{m^2 + n^2}$. Let $C H$ be the desired height. Expressing the area of triangle $A B C$ in two ways, we get

$$
\frac{1}{2} A B \cdot C H = \frac{1}{2} B C \cdot A C
$$

thus,

$$
C H = \frac{B C \cdot A C}{A B} = \frac{m \cdot m n}{m+n} = x^2 \cdot \frac{m n}{m+n} = \frac{(m+n)^2}{m^2 + n^2} \cdot \frac{m n}{m+n} = \frac{m n (m+n)}{m^2 + n^2}
$$

## Answer

An isosceles trapezoid with a $45^{\circ}$ angle at the base and a height of $\sqrt{2}$ is inscribed in a circle of radius 3. Find the area of the trapezoid.

## Solution

Let $A B C D$ be an isosceles trapezoid with bases $B C$ and $A D (B C < A D)$, $C H = \sqrt{2}$ be its height, $\angle A D C = 45^{\circ}$, and $R = 3$ be the radius of the circle. By the Law of Sines,

$$
A C = 2 R \sin \angle A D C = 6 \sin 45^{\circ} = 3 \sqrt{2}
$$

By the Pythagorean theorem,

$$
A H = \sqrt{A C^2 - C H^2} = \sqrt{18 - 2} = 4
$$

The projection $A H$ of the diagonal $A C$ of the isosceles trapezoid on the base $A D$ is equal to the half-sum of the bases, therefore,

$$
S_{A B C D} = A H \cdot C H = 4 \sqrt{2}
$$

## Answer

$4 \sqrt{2}$.