Auxiliary similar triangles [ Angle between tangent and chord Circle inscribed in an angle

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

A circle inscribed in an angle with vertex $O$ touches its sides at points $A$ and $B$. Ray $O X$ intersects this circle at points $C$ and $D$, such that

$O C=C D=1$. If $M$ is the point of intersection of ray $O X$ and segment $A B$, then what is the length of segment $O M$?
The first method. From the theorem about the angle between a tangent and a chord, it follows that $\angle O B C=\angle B D O$, hence, triangles $O B C$ and $O D B$ are similar by two angles, so $O B: O D=O C: O B$, from which $O B^{2}=O C \cdot O D=2, B C$ : $B D=O B: O D=\sqrt{2}: 2$.

Similarly, $A C: A D=\sqrt{2}: 2$.

Triangles $C M B$ and $A M D$ are similar by two angles, so $C M: A M=B C: A D$. Similarly, $A M: M D=A C$ : $B D$.

Multiplying the last two equalities, we get $\frac{C M}{M D}=\frac{B C}{A D} \cdot \frac{A C}{B D}=\frac{B C}{B D} \cdot \frac{A C}{A D}=\left(\frac{\sqrt{2}}{2}\right)^{2}=\frac{1}{2}$, hence, $C M=1 / 3$. Therefore, $O M=4 / 3$.

The second method. By the tangent-secant theorem, $O B^{2}=O C \cdot O D=2$.

Let $P$ be the projection of the center $Q$ of the circle onto the chord $C D, N$ be the point of intersection of $O Q$ and $A B$. Right triangles $O N M$ and $O P Q$ are similar, so

$O M: O Q=O N: O P$. The segment $B N$ is the altitude of the right triangle $O B Q$ dropped to the hypotenuse, so $O B^{2}=O Q \cdot O N$, hence, $O M=\frac{O Q \cdot O N}{O P}=\frac{O B^{2}}{O P}=\frac{2}{1+\frac{1}{2}}=\frac{4}{3}$.

Answer

## $4 / 3$