Concerning the Homothety of Tangent Circles helps to solve the problem, $\quad]$

In a right triangle $ABC$, angle $C$ is a right angle, and side $CA=4$. A point $D$ is taken on the leg $BC$, such that $CD=1$. A circle with radius $\frac{\sqrt{5}}{2}$ passes through points $C$ and $D$ and is tangent at point $C$ to the circumcircle of triangle $ABC$. Find the area of triangle $ABC$.
Since point $D$ lies inside the circumcircle of triangle $A B C$, the given circles touch internally. If $M$ is the point of intersection of the first circle with the leg $A C$ other than $C$, then $M D$ is the diameter of this circle,

$$
M C=\sqrt{M C D^{2}-C D^{2}}=\sqrt{5-1}=2
$$

Let $Q$ and $O$ be the midpoints of the diameters $M D$ and $A B$ of the given circles. Since $\angle C M D=\angle O C A=\angle B A C$, triangles $M C D$ and $A C B$ are similar with a ratio of $\frac{M I C}{C A}=\frac{1}{2}$. Therefore,

$$
B C=2 C D=2, S_{\triangle A B C}=\frac{1}{2} B C \cdot A C=4
$$

## Answer

4.