B2. Let $A B C$ be an isosceles triangle with the vertex at $C$, where $\angle A C B < 90^{\circ}$. Let $X$ be a point on side $A C$ different from $C$, and $Y$ be a point on side $B C$ different from $C$. Let $D$ be a point such that line $D X$ is parallel to line $A B$, and line $A C$ is the internal angle bisector of $\angle B A D$. Similarly, let $E$ be a point such that line $E Y$ is parallel to line $A B$, and line $B C$ is the internal angle bisector of $\angle E B A$. Suppose there exists a point $T$ on side $A B$ such that $|X T| = |X D|$ and $|Y T| = |Y E|$. Express the value of the expression $|A X| + |B Y|$ in terms of the lengths of the sides of triangle $A B C$.
B2.

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Let $\angle B A C = \angle C B A = \alpha$. Since line $A C$ is the angle bisector of $\angle B A D$, we have $\angle X A D = \alpha$. Due to the parallelism of lines $X D$ and $A B$, we also have $\angle D X A = \alpha$. Therefore, triangle $A X D$ is isosceles with the vertex at $D$, so $|X D| = |A D| = |T X|$. Since lines $A B$ and $D X$ are parallel, it follows that quadrilateral $A T X D$ is either a rhombus or an isosceles trapezoid. Similarly, we conclude that quadrilateral $T B E Y$ is either a rhombus or an isosceles trapezoid. We will show that both quadrilaterals are rhombuses.

By assumption, $\angle A C B = 180^{\circ} - 2 \alpha - 45^{\circ}$. If quadrilateral $A T X D$ is a rhombus, then $\angle X T A = 180^{\circ} - \angle T A D = 180^{\circ} - 2 \alpha$, and if it is an isosceles trapezoid, then $\angle X T A = \angle T A D = 2 \alpha$. Similarly, $\angle B T Y = 180^{\circ} - 2 \alpha$ if $T B E Y$ is a rhombus, and $\angle B T Y = 2 \alpha$ if $T B E Y$ is an isosceles trapezoid. Since points $X$ and $Y$ are different from $C$ and lie on the respective sides of the triangle, $\angle B T Y + \angle X T A + 4 \cdot 45^{\circ} = 180^{\circ}$. The combinations $\left(180^{\circ} - 2 \alpha\right) + 2 \alpha = 180^{\circ}$ and $2 \alpha + \left(180^{\circ} - 2 \alpha\right) = 180^{\circ}$ do not satisfy this condition, so the only possible combination is $\angle B T Y = \angle X T A = 180^{\circ} - 2 \alpha$. Therefore, quadrilaterals $A T X D$ and $T B E Y$ are rhombuses.

From this, it follows that triangles $A T X$ and $Y T B$ are isosceles with the vertex at $T$ and are similar to triangle $B C A$, as they share the angle at the base. Thus,

$$
\frac{|A X|}{|A B|} = \frac{|A T|}{|B C|} \quad \text{and} \quad \frac{|B Y|}{|A B|} = \frac{|B T|}{|A C|} = \frac{|B T|}{|B C|}
$$

Using these equalities, we express

$$
|A X| + |B Y| = \frac{|A B| \cdot |A T|}{|B C|} + \frac{|A B| \cdot |B T|}{|B C|} = \frac{|A B| \cdot (|A T| + |B T|)}{|B C|} = \frac{|A B|^2}{|B C|}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-32.jpg?height=57&width=1742&top_left_y=740&top_left_x=174)

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Determination that $A T X D$ or $T B E Y$ is a rhombus or an isosceles trapezoid...................... 1 point

Justification that both $A T X D$ and $T B E Y$ must be rhombuses ...............................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_21f8af0300261b15ee68g-32.jpg?height=54&width=1742&top_left_y=944&top_left_x=174)

Final calculation that $|A X| + |B Y| = \frac{|A B|^2}{|A C|} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .1$ point