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

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

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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.

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