[ Diameter, main properties

[ Theorem of tangent and secant; the product of the entire secant and its external part In triangle $A B C$, it is known that $A B=\sqrt{14}$ and $B C=2$. A circle is drawn through point $B$, through the midpoint $D$ of segment $B C$, through point $E$ on segment $A B$, and is tangent to side $A C$. Find the ratio in which this circle divides segment $A B$, if $D E$ is the diameter of this circle.
Apply the tangent-secant theorem.

## Solution

Since $E D$ is the diameter of the circle, then $\angle B=90^{\circ}$. Therefore,

$$
A C=\sqrt{B C^{2}+A B^{2}}=3 \sqrt{2}
$$

Let $P$ be the point of tangency of the circle with the hypotenuse $A C$. Then

$$
\begin{gathered}
P C^{2}=C B \cdot C D=2, P C=\sqrt{2}, A P=3 \sqrt{2}-\sqrt{2}=2 \sqrt{2} \\
A B \cdot A E=A P^{2}, \text { or } \sqrt{14}(\sqrt{14}-B E)=8
\end{gathered}
$$

From this, we find that

$$
B E=\frac{3 \sqrt{14}}{7}, A E=\sqrt{14}-\frac{3 \sqrt{14}}{7}=\frac{4 \sqrt{14}}{7}, \frac{A E}{B E}=\frac{4}{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-08.jpg?height=372&width=640&top_left_y=2075&top_left_x=706)

## Answer

$4: 3$.

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