# Problem 1. (2 points)

$A B C$ is an equilateral triangle with a side length of 10. On side $A B$, a point $D$ is taken; on side $A C$, a point $E$ is taken; on side $B C$, points $F$ and $G$ are taken such that triangles $A D E$, $B D G$, and $C E F$ are also equilateral. $A D=3$. Find $F G$.
# Answer: 4

## Solution:

$B D=A B-A D=10-3=7$, so all sides of triangle $B D G$ are equal to 7.

$C E=A C-A E=A C-A D=10-3=7$, so all sides of triangle $C E F$ are also equal to 7.

Thus, $B G+C F=14$. This is greater than the length of segment $B C=10$, so $G F=14-10=4$.