B2. Let $\mathcal{K}$ be a circle with diameter $A B$. Points $C$ and $D$ lie on the circle $\mathcal{K}$ on the same side of the line $A B$, such that the line $B D$ is the angle bisector of $\angle C B A$. The chords $A C$ and $B D$ intersect at point $E$. What is the length of the segment $D E$, if $|A E|=169$ and $|C E|=119$?
B2.

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-13.jpg?height=466&width=494&top_left_y=224&top_left_x=792)

Since line $B E$ is the angle bisector of $\angle C B A$, by the Angle Bisector Theorem of a triangle, $\frac{|A B|}{|B C|}=\frac{|A E|}{|C E|}=\frac{169}{119}$ or $|A B|=\frac{169}{119}|B C|$. By Thales' Theorem on the angle in a semicircle, $\angle A C B=\angle A D B=90^{\circ}$, so by the Pythagorean Theorem, $|A B|^{2}=|A C|^{2}+|B C|^{2}$ or $\frac{169^{2}}{119^{2}}|B C|^{2}=(169+119)^{2}+|B C|^{2}$. From this, we express $|B C|^{2}=\frac{119^{2}(169+119)^{2}}{169^{2}-119^{2}}=\frac{119^{2}(169+119)}{169-119}=\frac{119^{2} \cdot 288}{50}=\frac{119^{2} \cdot 144}{25}$. By the Pythagorean Theorem in triangle $B C E$, $|B E|=\sqrt{|C E|^{2}+|B C|^{2}}=\sqrt{119^{2}+\frac{119^{2} \cdot 144}{25}}=119 \sqrt{\frac{169}{25}}=119 \cdot \frac{13}{5}$. Since triangles $A D E$ and $B C E$ are similar, $\frac{|D E|}{|A E|}=\frac{|C E|}{|B E|}$, from which we calculate $|D E|=\frac{|A E| \cdot|C E|}{|B E|}=\frac{169 \cdot 119}{119 \cdot \frac{13}{5}}=5 \cdot 13=65$.

Note. The equality $\frac{|A B|^{5}}{|B C|}=\frac{|A E|}{|C E|}$ from the first part of the solution can also be proven using the Law of Sines. Let $\angle C B A=\beta$ and $\angle B E C=\varphi$. By the Law of Sines in triangle $B C E$, $\frac{|C E|}{|B C|}=\frac{\sin \frac{\beta}{2}}{\sin \varphi}$, and by the Law of Sines in triangle $B E A$, $\frac{|A E|}{|A B|}=\frac{\sin \frac{\beta}{2}}{\sin (\pi-\varphi)}$. Since $\sin (\pi-\varphi)=\sin \varphi$, it follows that $\frac{|A E|}{|A B|}=\frac{|C E|}{|B C|}$ or $\frac{|A B|}{|B C|}=\frac{|A E|}{|C E|}$.

2nd method. Let $F$ be the orthogonal projection of point $E$ onto segment $A B$. Since line $B E$ is the angle bisector of $\angle C B A$, triangles $B C E$ and $B F E$ are congruent, as they match in two angles and the side opposite the larger angle. Thus, $|B F|=|B C|$ and $|E F|=|C E|$. By Thales' Theorem on the angle in a semicircle, $\angle A C B=\angle A D B=90^{\circ}$, so by the Pythagorean Theorem,

$$
|B C|^{2}+|A C|^{2}=|A B|^{2}=(|A F|+|B F|)^{2}=(|A F|+|B C|)^{2}=|A F|^{2}+2 \cdot|A F| \cdot|B C|+|B C|^{2}
$$

From this, we express $|B C|=\frac{|A C|^{2}-|A F|^{2}}{2|A F|}$. By the Pythagorean Theorem for triangle $A F E$, $|A F|^{2}=|A E|^{2}-|E F|^{2}=|A E|^{2}-|C E|^{2}=169^{2}-119^{2}$, so

$$
\begin{aligned}
|B C| & =\frac{(169+119)^{2}-\left(169^{2}-119^{2}\right)}{2 \sqrt{169^{2}-119^{2}}}=\frac{2 \cdot 119 \cdot(169+119)}{2 \sqrt{(169+119)(169-119)}} \\
& =119 \cdot \sqrt{\frac{169+119}{169-119}}=119 \cdot \sqrt{\frac{288}{50}}=119 \cdot \sqrt{\frac{144}{25}}=119 \cdot \frac{12}{5}
\end{aligned}
$$

By the Pythagorean Theorem for triangle $B C E$, $|B E|=\sqrt{|B C|^{2}+|C E|^{2}}=\sqrt{119^{2} \cdot \frac{144}{25}+119^{2}}=119 \cdot \frac{13}{5}$. Since triangles $A D E$ and $B C E$ are similar, $\frac{|D E|}{|A E|}=\frac{|C E|}{|B E|}$, from which we calculate $|D E|=\frac{|A E| \cdot|C E|}{|B E|}=\frac{169 \cdot 119}{119 \cdot \frac{13}{5}}=5 \cdot 13=65$.

## 1st method

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-13.jpg?height=265&width=1762&top_left_y=2483&top_left_x=158)

$\frac{|A B|}{|B C|}=\frac{|A E|}{|C E|}$ 1 point

$|A B|=\frac{169}{119}|B C|$ 1 point

$\frac{169^{2}}{119^{2}}|B C|^{2}=(169+119)^{2}+|B C|^{2}$ 1 point

$|B C|^{2}=\frac{119^{2}(169+119)^{2}}{169^{2}-119^{2}}=\frac{119^{2} \cdot 144}{25}$. 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-14.jpg?height=768&width=1759&top_left_y=221&top_left_x=160)