11.5 The tangents to the circumscribed circle of the right triangle $A B C$ ( $\angle C=$ $90^{\circ}$ ) at points $A$ and $C$ intersect at point $T$. The rays $A B$ and $T C$ intersect at point $S$. It is known that the areas of triangles $\triangle A C T$ and $\triangle B C S$ are equal. Find the ratio of the areas of triangles $\triangle A B C$ and $\triangle A T S$.
Solution: Let point $M$ be the midpoint of the hypotenuse $AB$, and point $N$ be the midpoint of the leg $AC$. Note that points $M, N$, and $T$ lie on the same line - the perpendicular bisector of segment $AC$. Let the area of triangle $TNC$ be $S_{1}$, and the area of triangle $AMN$ be $S_{2}$. Then $S(\triangle ANT)=S_{1}, S(\triangle SBC)=2S_{1}, S(\triangle MNC)=S_{2}, S(\Delta BMC)=2S_{2}$. Since $S(\Delta TMC)=S_{1}+S_{2}, S(\Delta SMC)=2S_{1}+2S_{2}$, and these triangles have a common height, their bases are in the ratio $SC: TC=2: 1$. Then, by Thales' theorem and again using triangles with a common height

$$
\frac{SC}{TC}=\frac{SB}{BM}=\frac{S(\Delta SBC)}{S(\Delta BMC)}=\frac{S_{1}}{S_{2}}
$$

from which we get $S_{1}=2S_{2}$. Therefore, $S(\triangle ABC): S(\triangle ATS)=4S_{2}:(4S_{1}+4S_{2})=1: 3$.

## Criteria:

- The ratio $SC: TC$ is found - 3 points.