[Theorem of the tangent and secant; the product of the entire secant and its external part [Pythagorean Theorem (direct and inverse).

One of the two lines passing through point $M$ touches the circle at point $C$, while the second intersects this circle at points $A$ and $B$, with $A$ being the midpoint of segment $B M$. It is known that $M C=2$ and $\angle B M C=45^{\circ}$. Find the radius of the circle.

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Prove that triangle $B M C$ is a right triangle.

## Solution

Let $A M=A B=x$. By the tangent-secant theorem, $B M \cdot A M=M C^{2}$, or $2 x^{2}=4$, from which $x=\sqrt{2}$.

In triangle $B M C$, the sides $M C=2, B M=2 x=2 \sqrt{2}$ and the angle between them $\angle B M C=45^{\circ}$ are known. Consider a right triangle with a leg equal to 2 and an acute angle equal to $45^{\circ}$. Its hypotenuse is $2 \sqrt{2}$, so this triangle is congruent to triangle $B M C$ by two sides and the included angle. Therefore, triangle $B M C$ is a right triangle, $\angle B C M=90^{\circ}$. Then $B C$ is the diameter of the circle and $B C=M C=2$.

## Answer

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