1. (2 points) In trapezoid $A B C D$ with bases $A D=16$ and $B C=10$, the circles constructed on sides $A B, B C$ and $C D$ as diameters intersect at one point. The length of diagonal $A C$ is 10. Find the length of $B D$.
Answer: 24

## Solution:

Let the intersection point of the three circles be $O$. Then, since the circles are constructed on the sides of the trapezoid $AB, BC$, and $CD$ as diameters, the angles $\angle AOB, \angle BOC$, and $\angle COD$ are right angles. Therefore, points $A, O, C$ lie on the same line and points $B, O, D$ lie on the same line, meaning that $O$ is the intersection point of the diagonals of the trapezoid.

Let the midpoints of sides $AB, BC$, and $CD$ be $K, L$, and $M$ respectively. Since $KL$ and $LM$ are the midlines in triangles $ABC$ and $BCD$, $KL \parallel AC$ and $LM \parallel BD$. Thus, $KL \perp LM$, meaning that triangle $KLM$ is a right triangle. Moreover, $KL = \frac{AC}{2} = 5$, and $KM = \frac{AD + BC}{2} = 13$, so by the Pythagorean theorem, $LM = 12$. Therefore, $BD = 2LM = 24$.

Instead of reasoning with midlines, we can consider triangles $BOC$ and $AOD$, which are similar with a ratio of $10:16 = 5:8$. Since $AC = 10$, then $OC = \frac{5}{13}AC = \frac{50}{13}$. By the Pythagorean theorem in triangle $BOC$, we find $OB = \sqrt{10^2 - \frac{2500}{169}} = \frac{120}{13}$. Therefore, $BD = \frac{13}{5}OB = \frac{13}{5} \cdot \frac{120}{13} = 24$.