[ Auxiliary similar triangles ]

## $[\quad$ The Law of Sines

A circle is drawn through vertices $A$ and $B$ of triangle $ABC$, intersecting sides $BC$ and $AC$ at points $D$ and $E$ respectively. The area of triangle $CDE$ is seven times smaller than the area of quadrilateral $ABDE$. Find the chord $DE$ and the radius of the circle, given that $AB=4$ and $\angle C=45^{\circ}$.
Prove that triangles $C D E$ and $C A B$ are similar with a similarity coefficient of $\frac{1}{\sqrt{8}}$. Apply the Law of Sines to triangle $A D C$.

## Solution

Triangles $C D E$ and $C A B$ are similar by two angles.

By the given condition, $S_{C D E}=1 / 8 S_{C A B}$. Therefore, the similarity coefficient of triangles $C D E$ and $C A B$ is $\frac{1}{\sqrt{8}}$.

Thus, $E D=\frac{A B}{\sqrt{8}}=\sqrt{2}$.

Let $\angle C A D=\alpha$. Then $\angle A D C=135^{\circ}-\alpha$.

By the Law of Sines, $\frac{\sin \angle C A D}{\sin \angle A D C}=\frac{C D}{A C}=\frac{1}{\sqrt{8}}$, or $\frac{\sin \alpha}{\sin \left(135^{\circ}-\alpha\right)}=\frac{1}{\sqrt{8}}, 4 \sin \alpha=\sin \alpha+\cos \alpha$, from which $\operatorname{ctg} \alpha=3$. Therefore, $\sin \alpha=\frac{1}{\sqrt{10}}$.

Let $R$ be the desired radius. Then $R=\frac{D E}{2 \sin \angle E A D}=\frac{\sqrt{2}}{2 \sin \alpha}=\sqrt{5}$.

## Answer

$\sqrt{2}, \sqrt{5}$.