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


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 $\gamma=60^\circ$.
