B5. In a triangle $A B C$, angle $A$ is a right angle. On line segment $A B$ lies a point $D$ such that angles $A C D$ and $B C D$ are equal. It is given that $|A D|=2$ and $|B D|=3$.

What is the length of line segment $C D$?

## C-problems

![](https://cdn.mathpix.com/cropped/2024_04_17_fd6be77049c724ba1866g-2.jpg?height=434&width=394&top_left_y=705&top_left_x=1439)

For the C-problems, not only the answer is important; there should also be a reasoning that shows your answer is correct. Each correctly worked out C-problem is worth 10 points. Partial solutions can also earn points. Therefore, write everything clearly and submit your scratch paper as well.

NOTE: Solve each C-problem on a separate sheet of paper and also submit the corresponding scratch paper per problem.
B5. $2 \sqrt{6}$

We draw a perpendicular line from $D$ to side $B C$; the foot of the perpendicular we call $E$. It holds that $\angle A C D = \angle E C D$. Furthermore, we see the right angles $\angle D A C = \angle D E C$. The triangles $A C D$ and $E C D$ now share the side $C D$ and two equal angles, so the triangles $A C D$ and $E C D$ are mirror images of each other in $C D$. This implies that $|D E| = |D A| = 2$ and $|A C| = |E C|$.

For the rest, we will apply the Pythagorean theorem three times. First, we apply it in triangle $D E B$, which gives $2^{2} + |E B|^{2} = 3^{2}$, so

![](https://cdn.mathpix.com/cropped/2024_04_17_123a6e1baa7323ff717cg-2.jpg?height=434&width=397&top_left_y=1362&top_left_x=1435)
$|E B| = \sqrt{9 - 4} = \sqrt{5}$. Let $x = |A C| = |E C|$. Now we apply the Pythagorean theorem in triangle $A B C$, which gives

$$
x^{2} + (2 + 3)^{2} = (\sqrt{5} + x)^{2} \quad \text{so} \quad x^{2} + 25 = 5 + 2 \sqrt{5} x + x^{2} \quad \text{so} \quad 20 = 2 \sqrt{5} x.
$$

From this, it follows that $x = 2 \sqrt{5}$. The last time we apply the Pythagorean theorem is in triangle $C D E$: this gives $2^{2} + (2 \sqrt{5})^{2} = |C D|^{2}$ and thus $|C D| = \sqrt{24} = 2 \sqrt{6}$.

## C-problems