1. (2 points) In trapezoid $A B C D$ with bases $A D=20$ and $B C=14$, 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 16. Find the length of $B D$.
Answer: 30

## Solution:

Let the intersection point of the three circles be $O$. Then, since the circles are constructed on the sides $AB$, $BC$, and $CD$ of the trapezoid as diameters, the angles $\angle AOB$, $\angle BOC$, and $\angle COD$ are right angles. Therefore, points $A$, $O$, and $C$ lie on the same line, and points $B$, $O$, and $D$ 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, $KL = \frac{AC}{2} = 8$, and $KM = \frac{AD + BC}{2} = 17$, so by the Pythagorean theorem, $LM = 15$. Therefore, $BD = 2LM = 30$.

Instead of reasoning with midlines, we can consider triangles $BOC$ and $AOD$, which are similar with a ratio of $14:20 = 7:10$. Since $AC = 16$, then $OC = \frac{7}{17} AC = \frac{112}{17}$. By the Pythagorean theorem in triangle $BOC$, we find $OB = \sqrt{14^2 - \frac{112^2}{17^2}} = \frac{210}{17}$. Therefore, $BD = \frac{17}{7} OB = \frac{17}{7} \cdot \frac{210}{17} = 30$.