In triangle ABC, ∠ACB = 60. AD and BE are angle bisectors. Prove AE + BD = AB. Here is a diagram if it will be helpful:



Pick $G$ on $AB$ such that $AG=AE$. It follows that the angles $\angle AFG$ and $\angle AFE$ are equal. They are labelled $\alpha$ in the figure. Other angles $\beta$ and $\mu$ are also labelled. Now, $\angle ACB=60$ implies that $\alpha+\beta=120$: 
$$
\alpha+\beta=180-(\angle FAB+\angle FBA)=180-\frac{1}{2}(\angle CAB+\angle CBA)=180-\frac{1}{2}(180-60).
$$
But we also have $2\alpha+\beta=180$ and so $\alpha=\beta=\mu=60$. The triangles $\bigtriangleup GBF$ and $ \bigtriangleup DBF$ are congruent and the result follows.
