[ Angles subtended by equal arcs and equal chords]

## Law of Sines

The lengths of three sides of a cyclic quadrilateral inscribed in a circle of radius $2 \sqrt{2}$ are equal and each is 2. Find the fourth side.
Prove that the given quadrilateral is an isosceles trapezoid. Then apply the Law of Sines.

## Solution

Let the quadrilateral $A B C D$ be inscribed in a circle with radius $R=2 \sqrt{2}$, and $A B=B C=C D=2$. Let $\angle C A D=\alpha$. Since equal chords subtend equal arcs, and inscribed angles subtending equal chords are equal, $\angle A C B=\angle C A D$. Therefore, $B C \| A D$. Hence, $A B C D$ is an isosceles trapezoid.

From the Law of Sines, it follows that

$$
\sin \alpha=\sin \angle C A D=\frac{C D}{2 R}=\frac{2}{4 \sqrt{2}}=\frac{1}{2 \sqrt{2}}
$$

Since $\angle B A C=\angle C A D=\alpha$, then $\angle B A D=2 \alpha$

Let $P$ be the projection of point $B$ onto the larger base $A D$ of the trapezoid $A B C D$. From the right triangle $A B P$, we find that

$$
A P=A B \cos 2 \alpha=2\left(1-2 \sin ^{2} \alpha\right)=2\left(1-2 \cdot \frac{1}{8}\right)=\frac{3}{2}
$$

By the property of an isosceles trapezoid, $A P=\frac{A D-B C}{2}$. From this, we find that

$$
A D=2 A P+B C=2 \cdot \frac{3}{2}+2=3+2=5
$$

## Answer

5. 

## Problem