[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part] [The ratio of the areas of triangles with a common base or a common height

Two non-intersecting circles are inscribed in an angle $B$. The circle with the larger radius touches the sides of this angle at points $A$ and $C$, and the smaller circle touches the sides at points $A_{1}$ and $C_{1}$ (points $A, A_{1}$ and $C, C_{1}$ lie on different sides of angle $B$). The line $A C_{1}$ intersects the larger and smaller circles at points $E$ and $F$ respectively. Find the ratio of the areas of triangles $A B C_{1}$ and $A_{1} B C_{1}$, if $A_{1} B=2, E F=1$, and the length of $A E$ is the arithmetic mean of the lengths of $B C_{1}$ and $E F$.
Using the theorem of the tangent and the secant, prove that $A E=F C_{1}$.

## Solution

Since $B C_{1}=A_{1} B=2$, then

$$
A E=\frac{B C_{1}+E F}{2}=\frac{2+1}{2}=\frac{3}{2}
$$

From the theorem of the equality of segments of tangents drawn from a point to a circle, it follows that $C C_{1}=$ $A A_{1}$.

From point $A$, a tangent $A A_{1}$ and a secant $A F C_{1}$ are drawn to the smaller circle. By the theorem of the tangent and the secant, $A F \cdot A C_{1}=A A_{1}^{2}$.

From point $C_{1}$, a tangent $C_{1} C$ and a secant $C_{1} E A$ are drawn to the larger circle. Therefore, $C_{1} E \cdot A C_{1}=C_{1} C^{2}$.

Since $C C_{1}=A A_{1}$, then $A F \cdot A C_{1}=C_{1} E \cdot A C_{1}$. Therefore, $A F=C_{1} E$. Hence,

$$
C_{1} F=A E=\frac{3}{2}, A A_{1}=\sqrt{A C_{1} \cdot A F}=\sqrt{\left(\frac{3}{2}+1+\frac{3}{2}\right)\left(\frac{3}{2}+1\right)}=\sqrt{10}
$$

Therefore,

$$
\frac{S_{\triangle A B C_{1}}}{S_{\triangle A_{1} B C_{1}}}=\frac{A B}{A_{1} B}=\frac{\sqrt{10}+2}{2}=\frac{\sqrt{10}}{2}+1=\sqrt{\frac{5}{2}}+1 .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-43.jpg?height=528&width=866&top_left_y=-1&top_left_x=596)

## Answer

$1+\sqrt{\frac{5}{2}}$