2. Given the lengths of the legs $a=|B C|, b=|A C|$ of a right triangle $A B C$, where $a>b$. Let $D$ be the midpoint of the hypotenuse $A B$ and $E(E \neq C)$ the intersection of side $B C$ with the circumcircle of triangle $A D C$. Calculate the area of triangle $E A D$.


2. Let $c$ be the length of the hypotenuse $AB$, so $|AD|=|BD|=\frac{1}{2} c$. Quadrilateral $ADEC$ is cyclic and angle $ECA$ is a right angle, so the opposite angle $ADE$ is also a right angle (Fig. 1). Right triangles $ABC$ and $EDB$ share the angle at vertex $B$, so $\triangle ABC \sim \triangle EBD$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_04_17_76b6aaae36090a4ce582g-2.jpg?height=448&width=731&top_left_y=1478&top_left_x=662)

Fig. 1

$$
\frac{|ED|}{|BD|}=\frac{|AC|}{|BC|}, \quad \text { and thus } \quad|ED|=\frac{bc}{2a}
$$

The area of the right triangle $EAD$ is then (using the Pythagorean theorem)

$$
S=\frac{1}{2} \cdot|AD| \cdot|ED|=\frac{1}{2} \cdot \frac{c}{2} \cdot \frac{bc}{2a}=\frac{bc^2}{8a}=\frac{b(a^2+b^2)}{8a} .
$$

For a complete solution, award 6 points, of which 2 points for determining that angle $ADE$ is a right angle, 1 point for the similarity of triangles $ABC$ and $EBD$, 1 point for calculating the length of side $ED$, and 2 points for calculating the area of triangle $EAD$.