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 $30^{\circ}$, the area of triangle $AEC$ is $\sqrt{3} / 2$, 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-angled, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$, $A C=\frac{E C}{\sin 30^{\circ}}=2 E C, A E=E C \operatorname{tg} 60^{\circ}=\sqrt{3} E C ;$ $S_{A E C}=\frac{A E \cdot E C}{2}=\frac{\sqrt{3} E C^{2}}{2}=\frac{\sqrt{3}}{2} \Rightarrow E C=1$, $A C=2, A E=\sqrt{3} ;$
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: $\frac{D E}{\sin (\angle D C E)}=\frac{E C}{\sin 30^{\circ}}$

![](https://cdn.mathpix.com/cropped/2024_05_06_9cd2bbfc846d490b2ee8g-17.jpg?height=891&width=606&top_left_y=1502&top_left_x=1316)
$\Rightarrow \frac{\sqrt{2}}{\sin (\angle D C E)}=2 \Rightarrow \sin (\angle D C E)=\frac{\sqrt{2}}{2} \Rightarrow \angle D C E=45^{\circ} \Rightarrow \triangle E O C$ - isosceles right-angled triangle, $E O=E C=1$;
5) $\angle D A E=45^{\circ} \Rightarrow \triangle A B E$ - isosceles right-angled triangle, $B E=A E=\sqrt{3}$
6) $\triangle B E O$ - right-angled 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