[ Law of Sines

[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part

Given a triangle $A B C$. A circle of radius $R$ touches side $A C$ at point $M$ and side $B C$ at point $P$. Side $A B$ intersects this circle at points $K$ and $E$ (point $E$ lies on segment $B K$). Find $B E$, given that $B C=a, C M=$ $b<a, \angle K M E=\alpha$.

#
Apply the Tangent-Secant Theorem.

## Solution

$$
P C=C M=b, B P=B C-P C=a-b, K E=2 R \sin \alpha
$$

By the Tangent-Secant Theorem,

$$
B E \cdot(B E+K E)=B P^{2} \text {, or } B E^{2}+B E \cdot 2 R \sin \alpha=(a-b)^{2} \text {. }
$$

From this equation, we find that

$$
B E=\sqrt{R^{2} \sin ^{2} \alpha+(a-b)^{2}}-R \sin \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-46.jpg?height=472&width=652&top_left_y=627&top_left_x=703)

## Answer

$\sqrt{R^{2} \sin ^{2} \alpha+(a-b)^{2}}-R \sin \alpha$.