A triangle $$ABC$$ is turned about the vertex $$A$$ in the plane of the triangle into the position $$AB'C'$$. If $$\overline{AC}$$ bisects $$\overline{BB'}$$, prove that $$\overline{AB'}$$ bisects $$\overline{CC'}$$.

Let the rotation be through angle $$\theta$$, so $$AB' = AB$$, $$AC' = AC$$, $$\angle BAB' = \angle CAC' = \theta$$ and $$\angle B'AC' = \angle BAC$$. Triangle $$ABB'$$ is isosceles, so the line from $$A$$ through the midpoint of $$BB'$$ is the bisector of $$\angle BAB'$$; as $$AC$$ passes through that midpoint, $$\angle BAC = \angle CAB' = \frac{\theta}{2}$$. Then $$\angle B'AC' = \angle BAC = \frac{\theta}{2}$$ as well, so $$AB'$$ bisects $$\angle CAC'$$. In the isosceles triangle $$ACC'$$ the bisector of the apex angle also bisects the base, so $$AB'$$ bisects $$\overline{CC'}$$.