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Two circles with centers $O$ and $Q$, intersecting each other at points $A$ and $B$, intersect the bisector of angle $O A Q$ at points $C$ and $D$ respectively. Segments $A D$ and $O Q$ intersect at point $E$, and the areas of triangles $O A E$ and $Q A E$ are 18 and 42 respectively. Find the area of quadrilateral $O A Q D$ and the ratio $B C: B D$.
Triangles $D E Q$ and $A E O$ are similar, as are triangles $B C D$ and $A O Q$.

## Solution

Notice that $E Q: E O=S_{Q A E}: S_{O A E}=7: 3$.

Triangle $A Q D$ is isosceles, so $\angle O A D=\angle Q A D=\angle Q D A$. Therefore, $Q D \| O A$, which means triangles $D E Q$ and $A E O$ are similar with a similarity ratio of $E Q / E O=7 / 3$. Thus, $S_{D E Q}=49 / 9 S_{A E O}=98$, and $S_{O D E}=3 / 7 S_{Q D E}=42$.

Therefore, $S_{O A Q D}=S_{Q A E}+S_{O A E}+S_{O D E}+S_{Q D E}=42+18+42+98=200$.

$\angle A Q O=1 / 2 \angle A Q B=\angle A D B, \angle B C D=180^{\circ}-\angle A C B=1 / 2 \angle A O B=\angle A O Q$. From the proven angle equalities, the similarity of triangles $B C D$ and $A O Q$ follows. Therefore, $B C: B D=A O: A Q=E O: E Q=3: 7$.

## Answer

$200 ; 3: 7$.