7. On the side $AC$ of triangle $ABC$, a circle is constructed with $AC$ as its diameter, which intersects sides $AB$ and $BC$ at points $D$ and $E$ respectively. The angle $EDC$ is $30^{\circ}$, $AE = \sqrt{3}$, and the area of triangle $DBE$ is to the area of triangle $ABC$ as $1:2$. Find the length of segment $BO$, where $O$ is the intersection point of segments $AE$ and $CD$.
# Solution:

1) $\angle E D C=\angle E A C=30^{\circ}$ (inscribed angles subtending the same arc);
2) $A C$ - diameter of the circle $\Rightarrow \triangle A E C$ - right triangle, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$, $A C=\frac{A E}{\cos 30^{\circ}}=2, E C=A C \sin 30^{\circ}=1$

![](https://cdn.mathpix.com/cropped/2024_05_06_9cd2bbfc846d490b2ee8g-03.jpg?height=882&width=629&top_left_y=1775&top_left_x=1139)
3) $\angle A D C=90^{\circ}, \angle E D C=30^{\circ} \Rightarrow \angle B D E=60^{\circ} \Rightarrow \triangle D B E \approx \triangle C B A \Rightarrow \frac{D E}{A C}=\frac{B D}{B C}=\frac{B E}{A B}=k$, $k^{2}=\frac{S_{D B E}}{S_{A B C}}=\frac{1}{2} \Rightarrow k=\frac{1}{\sqrt{2}} \Rightarrow D E=\sqrt{2} ;$
4) $\triangle D E C$ - Law of Sines: $\frac{D E}{\sin (\angle D C E)}=\frac{E C}{\sin 30^{\circ}} \Rightarrow \frac{\sqrt{2}}{\sin (\angle D C E)}=2 \Rightarrow \sin (\angle D C E)=\frac{\sqrt{2}}{2}$ $\Rightarrow \angle D C E=45^{\circ} \Rightarrow \triangle E O C$ - isosceles right triangle, $E O=E C=1$;
5) $\angle D A E=45^{\circ} \Rightarrow \triangle A B E$ - isosceles right triangle, $B E=A E=\sqrt{3}$;
6) $\triangle B E O$ - right triangle $\Rightarrow B O^{2}=B E^{2}+E O^{2}$ (Pythagorean theorem) $\Rightarrow$ $B O^{2}=3+1=4 \Rightarrow B O=2$.

Answer: 2 .