Prove that the continuation of the common chord of the specified circles passes through the midpoint of $B C$.

## Solution

Let $D$ be the second point of intersection of the specified circles, and the line $A D$ intersects $B C$ at point $M$. Then, by the tangent-secant theorem,

$$
M B^{2}=M A \cdot M D=M C^{2}
$$

therefore $M B=M C$, i.e., $A M$ is the median of triangle $A B C, A M=3$. From the equation

$$
M A(M A-A D)=M B^{2}, \text { or } 3(3-A D)=4
$$

we find that $A D=\frac{5}{3}$.

Let $D$ be the second point of intersection of the specified circles, and the line $A D$ intersects $B C$ at point $M$. Then, by the tangent-secant theorem,

$$
M B^{2}=M A \cdot M D=M C^{2}
$$

therefore $M B=M C$, i.e., $A M$ is the median of triangle $A B C, A M=3$. From the equation

$$
M A(M A-A D)=M B^{2}, \text { or } 3(3-A D)=4
$$

we find that $A D=\frac{5}{3}$.

Let $D$ be the second point of intersection of the specified circles, and the line $A D$ intersects $B C$ at point $M$. Then, by the tangent-secant theorem,

$$
M B^{2}=M A \cdot M D=M C^{2}
$$

therefore $M B=M C$, i.e., $A M$ is the median of triangle $A B C, A M=3$. From the equation

$$
M A(M A-A D)=M B^{2}, \text { or } 3(3-A D)=4
$$

we find that $A D=\frac{5}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-32.jpg?height=431&width=626&top_left_y=1151&top_left_x=722)

Answer

$\frac{5}{3}$.
In triangle $ABC$, side $BC$ is equal to 4, and the median drawn to this side is equal to 3. Find the length of the common chord of two circles, each of which passes through point $A$ and is tangent to $BC$, with one touching $BC$ at point $B$ and the other at point $C$.