In triangle $A B C$ with perimeter $2 p$, the acute angle $B A C$ is $\alpha$. A circle with center at point $O$ touches side $B C$ and the extensions of sides $A B$ and $A C$ at points $K$ and $L$ respectively. Find the area of triangle $A O L$.

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The segments of tangents drawn from one point to a circle are equal to each other.

## Solution

Let $M$ be the point of tangency of the given circle with side $B C$. Then

$$
K B=B M, L C=C M, 2 p=A B+B C+A C=A K+A L
$$

and since $A K=A L$, then $A L=p$. Therefore,

$$
O L=\operatorname{ALtg} \frac{\alpha}{2}=p t g \frac{\alpha}{2}
$$

Thus,

$$
S_{\Delta \mathrm{AOL}}=\frac{1}{2} A L \cdot L O=\frac{1}{2} p^{2} \operatorname{tg} \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-43.jpg?height=432&width=960&top_left_y=2111&top_left_x=549)

Answer

$\frac{1}{2} p^{2} \operatorname{tg} \frac{\alpha}{2}$

Prove that the segment of the common external tangent to two circles, enclosed between the common internal tangents, is equal to the segment of the common internal tangent.

## Hint

The segments of tangents drawn from one point to a circle are equal to each other.

## Solution

Let $P$ and $Q$ be the endpoints of the first segment, $P Q=a ; x$ and $y$ be the segments of the tangents from point $P$ to the point of tangency with the first circle and from point $Q$ to the point of tangency with the second circle; $b$ be the segment of the common internal tangent.

The segments of tangents drawn from point $P$ to the second circle are equal to each other: $a+y=x+b$.

Similarly for point $Q: a+x=y+b$. Therefore, $a=b$ (and $x=y$).

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-44.jpg?height=417&width=778&top_left_y=703&top_left_x=640)

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