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

 A: The main idea is to express everything in terms of the triangle sides and then obtain something trivial (or at least easy-to-prove).
For this, just use the property that the angle bisector divides the opposite side in the ratio of the neighbour sides i.e. $\frac{AE}{EC}=\frac{AB}{BC}$.
Then (with the usual notations for triangle sides) you get
$AE=\frac{bc}{a+c}$ and $BD=\frac{ac}{b+c}$ and so we need to prove
$c=\frac{bc}{a+c}+\frac{ac}{b+c}$
which is (after expanding) equivalent to $a^2-ab+b^2=c^2$.
But this is just the Law of Cosines using $\gamma=60^\circ$.
A: 
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.
