3. (8 points) In a circle, chords $A B$ and $A C$ are drawn, with $A B=2, A C=1$, and $\angle C A B=120^{\circ}$. Find the length of the chord of the circle that bisects angle $C A B$.

Answer: 3
Solution. $A E=A C=1$, triangles $A C E$ and $B C D$ are equilateral. When rotated $60^{\circ}$ around point $C$, triangle $A B C$ transforms into triangle $E D C$. Therefore, $A B=E D=2$.

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