3. The lengths of the sides of triangle ABC are $\overline{\mathrm{AB}}=25 \mathrm{~cm}, \overline{\mathrm{BC}}=7 \mathrm{~cm}$, and $\overline{\mathrm{AC}}=24$. The angle bisector of the angle at vertex B intersects side AC at point D. The perpendicular bisector of segment $\mathrm{BD}$ intersects $\mathrm{BC}$ at $\mathrm{X}$, and $\mathrm{AB}$ at Y. Find the length of segment XY.
Solution. From the converse of the theorem of

![](https://cdn.mathpix.com/cropped/2024_06_05_fb56a3883e438b3946fbg-4.jpg?height=233&width=346&top_left_y=773&top_left_x=1132)

Fig. 6

The Pythagorean theorem implies that triangle ABC is a right triangle with a right angle at vertex C (Fig. 6). Since BD is the angle bisector of angle CBA, it follows that $\overline{\mathrm{BC}}: \overline{\mathrm{BA}}=\overline{\mathrm{CD}}: \overline{\mathrm{DA}}$. Therefore, $24=\overline{\mathrm{CD}}+\frac{25}{7} \overline{\mathrm{CD}}=\frac{32}{7} \overline{\mathrm{CD}}$, hence $\overline{\mathrm{CD}}=\frac{21}{4}$. Then $\overline{\mathrm{BD}}=\sqrt{49+\frac{441}{16}}=\frac{35}{4}$. Since triangles YBS and XBS are congruent, it follows that $\overline{X Y}=2 \overline{S Y}$. From the similarity of triangles YBS and DBC and from $\overline{\mathrm{BS}}=\overline{\mathrm{SD}}=\frac{\overline{\mathrm{BD}}}{2}=\frac{35}{8}$, we get $\overline{\mathrm{BS}}: \overline{\mathrm{BC}}=\overline{\mathrm{SY}}: \overline{\mathrm{CD}}, \overline{\mathrm{XY}}=\frac{21 \cdot 35}{4 \cdot 4 \cdot 7}=\frac{105}{16} \mathrm{~cm}$.