B2. Consider a circle with diameter $A B$. Point $C$ lies on the line segment $A B$ such that $2 \cdot|A C|=|B C|$. The points $D$ and $E$ lie on the circle so that $C D$ is perpendicular to $A B$ and $D E$ is also a diameter of the circle. Denote the areas of triangles $A B D$ and $C D E$ as $O(A B D)$ and $O(C D E)$. Determine the value of $\frac{O(A B D)}{O(C D E)}$.
B2. Let the center of the circle be $M$. Triangles $C D M$ and $C E M$ have equal areas. Indeed, the two triangles have bases of equal length $|D M|=|E M|$ and the same height. It follows that

$$
O(C D E)=2 \cdot O(C D M)
$$

We have $|A C|=\frac{1}{3}|A B|$ and $|A M|=\frac{1}{2}|A B|$, so $|C M|=|A M|-|A C|=$ $\frac{1}{2}|A B|-\frac{1}{3}|A B|=\frac{1}{6}|A B|$. Triangles $A B D$ and $C D M$ have the same height relative to the bases $A B$ and $C M$, and thus

$$
O(A B D)=6 \cdot O(C D M)
$$

Combining the two equations gives $\frac{O(A B D)}{O(C D E)}=\frac{6}{2}=3$.

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