5. (8 points) Chord $CD$ is parallel to the diameter $AB$ of a circle with center at point $O$ and radius $r$. The tangent to this circle at point $A$ intersects the lines $BC$ and $BD$ at points $E$ and $F$ respectively. Find $AE \cdot AF$.
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_e53c58234eff2beb6c25g-05.jpg?height=477&width=914&top_left_y=441&top_left_x=614)

Since in the inscribed quadrilateral $ABDC$, the opposite sides $AB$ and $CD$ are parallel, $ABDC$ is an isosceles trapezoid ($BD = AC$). Therefore, $\angle DBA = \angle BAC$. Since $AB$ is the diameter, $\angle BCA = 90^\circ$. And since $AE$ is a tangent, $\angle BAE = 90^\circ$. We get

$$
\angle BEA = 90^\circ - \angle EBA = 90^\circ - \angle CBA = 90^\circ - (90^\circ - \angle BAC) = \angle BAC
$$

Therefore,

$$
\angle FBA = \angle DBA = \angle BAC = \angle BEA
$$

Triangles $FBA$ and $BEA$ are similar by two angles ($\angle FBA = \angle BEA$ and $\angle FAB = \angle BAE = 90^\circ$).

$$
\frac{AB}{AE} = \frac{AF}{AB} \Longrightarrow AE \cdot AF = AB^2 = (2r)^2 = 4r^2
$$

Answers:

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