3. Given an equilateral triangle ABC. Point $D$ is chosen on the extension of side $A B$ beyond point $A$, point $E$ is on the extension of $B C$ beyond point $C$, and point $F$ is on the extension of $A C$ beyond point $C$ such that $C F=A D$ and $A C+E F=D E$. Find the angle BDE. (A. Kuznetsov)
Answer: 60 - . Solution: Complete triangle $A C E$ to parallelogram $A C E G$. Since $C F=A D$, $C E=A G$ and $\cdot F C E=\cdot D A G=60 \cdot$, triangles $D A G$ and $F C E$ are equal, from which $G D=E F$. Therefore, $D E=A C+E F=G E+G D$. This means that point $G$ lies on segment $D E$, and thus $D E \| A C$, from which $\cdot B D E=\cdot B A C=60$