An isosceles right triangle $ABC$ ($\angle B-$ right angle), with an area of $4+2\sqrt{2}$, is inscribed in a circle. Point $D$ lies on this circle, and the chord $BD$ equals 2. Find the chords $AD$ and $CD$.

#
Using the cosine theorem, formulate an equation with respect to $A D$.

## Solution

Let $a$ be the legs of the triangle. From the problem statement, we have

$$
\frac{a^{2}}{2}=4+2 \sqrt{2}
$$

Therefore, $a^{2}=8+4 \sqrt{2}$.

Let point $D$ lie on the arc $A B$ that does not contain point $C$. Since

$$
\angle A D B=180^{\circ}-\angle A C B=180^{\circ}-45^{\circ}=135^{\circ}
$$

by the cosine theorem,

$$
A B^{2}=A D^{2}+B D^{2}-2 A D \cdot B D \cos 135^{\circ}
$$

or

$$
8+4 \sqrt{2}=A D^{2}+4+A D \sqrt{2}
$$

From this, we find that

$$
A D=-\sqrt{6+4 \sqrt{2}}=-\sqrt{2}+2+\sqrt{2}=2
$$

Similarly, from triangle $B D C\left(\angle B D C=\angle B A C=45^{\circ}\right)$, we find that $D C=2(\sqrt{2}+1)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-44.jpg?height=469&width=514&top_left_y=2238&top_left_x=772)

## Answer

$2 ; 2(1+\sqrt{2})$ or $2(1+\sqrt{2}) ; 2$