On a circle of radius 5, circumscribed around an equilateral triangle, a point $D$ is taken. It is known that the distance from point $D$ to one of the vertices of the triangle is 9. Find the sum of the distances from point $D$ to the other two vertices of the triangle.
Let the distance from vertex $A$ of triangle $A B C$ to point $D$ be 9. Prove that point $D$ lies on the smaller arc $B C$ and $B D + C D = A D$.

## Solution

Let the distance from vertex $A$ of triangle $A B C$ to point $D$ be 9. Note that $B C = 5 \sqrt{3}$. Since

$A D = 9 > B C$, point $D$ lies on the smaller arc $B C$.

Construct a segment $A K$ on ray $A D$ equal to $B D$. Then triangle $A K C$ is congruent to triangle $B D C$ by two sides and the included angle

( $\angle K A C = \angle D A C = \angle D B C$ as inscribed angles subtending the same arc). Therefore, $C K = C D$, and since $\angle K D C = \angle A D C = \angle B = 60^{\circ}$, triangle $C K D$ is equilateral, meaning $D K = D C$. Since $A K = B D < B C < A D$, point $K$ lies on segment $A D$. Therefore, $B D + C D = A K + K D = A D = 9$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-47.jpg?height=783&width=778&top_left_y=0&top_left_x=640)

## Answer

9.