## Task A-1.4.

Let $\overline{A C}$ be the diameter of circle $k_{1}$ whose center is at point $B$. Circle $k_{2}$ touches the line $A C$ at point $B$ and circle $k_{1}$ at point $D$. The tangent from $A$ (different from $A C$) to circle $k_{2}$ touches this circle at point $E$ and intersects the line $B D$ at point $F$. Determine the ratio $|A F|:|A B|$.
## First solution.

Let $S$ be the center of circle $k_{2}$ and let $|B S|=|E S|=r$ be the radius of that circle.

![](https://cdn.mathpix.com/cropped/2024_05_30_659ee8c29a87e0ecc7d7g-04.jpg?height=765&width=699&top_left_y=934&top_left_x=610)

Since point $S$ is the midpoint of segment $\overline{B D}$, which is the radius of circle $k_{1}$, it follows that $|A B|=|B D|=2 r$.

We have $|A E|=|A B|=2 r$ because these are segments of the tangent.

Since the tangent $A B$ to circle $k_{2}$ is perpendicular to the radius $\overline{S B}$, triangle $A B F$ is a right triangle.

Similarly, since the tangent $A E$ to circle $k_{2}$ is perpendicular to the radius $\overline{S E}$ of circle $k_{2}$, triangle $S E F$ is a right triangle.

1 point

Right triangles $A B F$ and $S E F$ have the same angles, so they are similar. 2 points

Let $a=|F S|$. From the mentioned similarity, we conclude $|A F|=|F S| \cdot \frac{|A B|}{|E S|}=2|F S|=2 a$ 1 point

and conclude

$$
2 a-2 r=|A F|-|A E|=|E F|=\frac{1}{2} \cdot|B F|=\frac{1}{2}(r+a)
$$

From this, it follows that $3 a=5 r$, or $|A F|:|A B|=2 a: 2 r=5: 3$. 1 point