B2. Let $A B C$ be a right triangle with a right angle at $C$, for which $|B C|=a$ and $|A C|=b$. Let $D$ be a point that lies on the opposite side of the line $A C$ from point $B$, such that triangle $A C D$ is similar to triangle $A B C$. Let $E$ be a point on the line $C D$, such that the angle $\Varangle E B C$ is a right angle. Express the area of quadrilateral $A B E D$ in terms of $a$ and $b$.
B2. Due to the similarity of triangles $A C D$ and $A B C$, we have $\Varangle C A D = \Varangle B A C$. We observe that point $E$ lies on the opposite side of line $B C$ compared to point $A$, so $\Varangle B C E = 180^{\circ} - \Varangle A C B - \Varangle D C A = 90^{\circ} - \Varangle D C A = \Varangle C A D = \Varangle B A C$. Therefore, triangles $A B C$ and $C E B$ are also similar, as they have two equal angles. By the Pythagorean theorem, we have $|A B| = \sqrt{a^{2} + b^{2}}$. From the similarity of triangles $A B C$ and $A C D$, it follows that

$$
\frac{|A D|}{|A C|} = \frac{|A C|}{|A B|} \quad \text{and} \quad \frac{|D C|}{|A C|} = \frac{|C B|}{|A B|}
$$

From the first equation, we express $|A D| = \frac{|A C|^{2}}{|A B|} = \frac{b^{2}}{\sqrt{a^{2} + b^{2}}}$, and from the second, $|D C| = \frac{|A C| \cdot |C B|}{|A B|} = \frac{a b}{\sqrt{a^{2} + b^{2}}}$. Since triangles $A B C$ and $C E B$ are also similar, we have

$$
\frac{|B E|}{|B C|} = \frac{|C B|}{|C A|}
$$

from which we express $|B E| = \frac{|C B|^{2}}{|C A|} = \frac{a^{2}}{b}$. The area of quadrilateral $A B E D$ is therefore

$$
\begin{aligned}
p & = \frac{|A D| \cdot |D C|}{2} + \frac{|A C| \cdot |C B|}{2} + \frac{|C B| \cdot |B E|}{2} = \\
& = \frac{a b^{3}}{2(a^{2} + b^{2})} + \frac{a b}{2} + \frac{a^{3}}{2 b} = \frac{a b^{4} + a b^{2}(a^{2} + b^{2}) + a^{3}(a^{2} + b^{2})}{2 b(a^{2} + b^{2})} = \\
& = \frac{2 a b^{4} + 2 a^{3} b^{2} + a^{5}}{2 b(a^{2} + b^{2})}
\end{aligned}
$$

Clear and labeled sketch 1 point Well-justified conclusion that triangles $A B C$ and $C E B$ are similar 1 point Application of the similarity of triangles $A B C$ and $A C D$ and the expression of sides $|A D|$ and $|D C|$ 1 point

Application of the similarity of triangles $A B C$ and $C E B$ and the expression of side $|B E|$ 1 point Expression of the area of quadrilateral $A B E D$ as the sum of the areas of three right triangles 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_a0d1506e8e2028d6d427g-20.jpg?height=55&width=1576&top_left_y=2000&top_left_x=274)