Two circles with centers $O_{1}$ and $O_{2}$ intersect at points A and B. The first circle passes through the center of the second, and its chord $B D$ intersects the second circle at point $C$ and divides the arc $A C B$ in the ratio $A C: C B=n$. In what ratio does point $D$ divide the arc $A D B$?
Express the indicated arcs in terms of $n$.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the circles. Suppose the angular measures of the arcs $A C$ and $B C$ of the second circle are $n x$ and $x$. Then $\angle A B D=\frac{n x}{2}$ and the angular measure of the arc $A D$ of the first circle is $n x$.

From the isosceles triangle $A O^{2} B$, we find that

$$
\angle O_{2} A B=\frac{1}{2}\left(180^{\circ}-(n+1) x\right)
$$

Then the angular measure of the arc $A B$ of the first circle is four times that, i.e.,

$$
\cup A B=2\left(180^{\circ}-(n+1) x\right) \text {, }
$$

and the angular measure of the supplementary arc of the first circle is $2(n+1) x$. Therefore, the desired ratio is

$$
\frac{n x}{(2(n+1) x-n x)}=\frac{n}{n+2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-25.jpg?height=441&width=554&top_left_y=2152&top_left_x=752)

Answer

$\cup A D: \cup D B=n:(n+2)$.

Two circles touch each other internally at point $A$; $A B$ is the diameter of the larger circle. The chord $B K$ of the larger circle touches the smaller circle at point $C$. Prove that $A C$ is the bisector of triangle $A B K$.

## Hint

Prove that $C O \| A K(O-$ is the center of the smaller circle).

## Solution

Let $O$ be the center of the smaller circle. Since

$$
\angle B C O=\angle B K A=90^{\circ}
$$

then $C O \| A K$. Therefore, $\angle A C O=\angle K A O$, and since $O C=O A$, triangle $C O A$ is isosceles and $\angle A C O=\angle C A O$. Consequently, $\angle K A C=\angle C A O$, i.e., $A C$ is the bisector of triangle $A B K$.

The statement remains true if $A B$ is any chord of the larger circle.

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-26.jpg?height=408&width=512&top_left_y=928&top_left_x=773)

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