In a right triangle $A B C$ with legs $A B=3$ and $B C=4$, a circle is drawn through the midpoints of sides $A B$ and $A C$, touching the leg $B C$. Find the length of the segment of the hypotenuse $A C$ that lies inside this circle.
Apply the tangent-secant theorem.

## Solution

Let $M$ and $N$ be the midpoints of $A C$ and $A B$ respectively, and $K$ be the point of tangency. Then $M N$ is the midline of triangle $A B C$; the diameter of the circle passing through the point of tangency $K$ is perpendicular to $B C$, and therefore also to $M N$. Therefore,

$$
M N=\frac{1}{2} B C=2, C K=C B-K B=C B-\frac{1}{2} M N=4-1=3 .
$$

Let $T$ be the second point of intersection of the circle with the hypotenuse $A C$. Then

$$
C T \cdot C M=C K^{2}, \text { or }\left(\frac{5}{2}+M T\right) \frac{5}{2}=9
$$

From this, we find that $M T=\frac{11}{10}$.

Let $M$ and $N$ be the midpoints of $A C$ and $A B$ respectively, and $K$ be the point of tangency. Then $M N$ is the midline of triangle $A B C$; the diameter of the circle passing through the point of tangency $K$ is perpendicular to $B C$, and therefore also to $M N$. Therefore,

$$
M N=\frac{1}{2} B C=2, C K=C B-K B=C B-\frac{1}{2} M N=4-1=3
$$

Let $T$ be the second point of intersection of the circle with the hypotenuse $A C$. Then

$$
C T \cdot C M=C K^{2}, \text { or }\left(\frac{5}{2}+M T\right) \frac{5}{2}=9
$$

From this, we find that $M T=\frac{11}{10}$.

Let $M$ and $N$ be the midpoints of $A C$ and $A B$ respectively, and $K$ be the point of tangency. Then $M N$ is the midline of triangle $A B C$; the diameter of the circle passing through the point of tangency $K$ is perpendicular to $B C$, and therefore also to $M N$. Therefore,

$$
M N=\frac{1}{2} B C=2, C K=C B-K B=C B-\frac{1}{2} M N=4-1=3
$$

Let $T$ be the second point of intersection of the circle with the hypotenuse $A C$. Then

$$
C T \cdot C M=C K^{2}, \text { or }\left(\frac{5}{2}+M T\right) \frac{5}{2}=9
$$

From this, we find that $M T=\frac{11}{10}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-06.jpg?height=403&width=688&top_left_y=1857&top_left_x=685)

## Answer

$\frac{11}{10}$.