Intersecting Circles $\quad]$ [Ratio of Areas of Similar Triangles]

Two circles with radii 3 and 4, the distance between their centers being 5, intersect at points $A$ and $B$. A line through point $B$ intersects the circles at points $C$ and $D$, such that $C D=8$ and point $B$ lies between points $C$ and $D$. Find the area of triangle $A C D$.
Let $O_{1}$ and $O_{2}$ be the centers of the circles. Prove that triangle $A C D$ is similar to triangle $B O_{1} O_{2}$.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the smaller and larger circles, respectively, and let point $C$ be located on the smaller circle. Then

$$
\angle D C A=\angle B C A=\frac{1}{2} \cup A B=\angle B O_{1} O_{2}
$$

Similarly, $\angle C D A=\angle B O_{2} O_{1}$. Therefore, triangle $A C D$ is similar to triangle $B O_{1} O_{2}$ with a similarity ratio of $\frac{C D}{\mathrm{O}_{1} \mathrm{O}_{2}}=\frac{8}{5}$.

Triangle $\mathrm{BO}_{1} \mathrm{O}_{2}$ is a right triangle, since

$$
O_{1} O^{2}{ }_{2}=5^{2}=3^{2}+4^{2}=O_{1} B^{2}+O_{2} B^{2} .
$$

Therefore,

$$
S_{\triangle \mathrm{BCD}}=\frac{64}{25} S_{\Delta \mathrm{BO}_{1} \mathrm{O}_{2}}=\frac{384}{25}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-16.jpg?height=421&width=706&top_left_y=1&top_left_x=676)

## Answer

$\frac{384}{25}$.