Prove that BF + CE = BC In $\triangle ABC$ , $\angle A = 60$
$~BE$  is the bisector of $\angle B$ and $~A-E-C.$
$~CF$  is the bisector of $\angle C$ and $~A-F-B$
Prove that $~BF + CE = BC$

let $AB = c, BC = a, CA = b$. by bisector, we have $\frac{CE}{AE} = \frac{a}{c}, \frac{BF}{AF} = \frac{a}{b}$.
by equal ratios, we have $\frac{CE}{AE+CE} = \frac{a}{a+c}, \frac{BF}{AF+BF} = \frac{a}{a+b}$, as $\frac{CE}{b} = \frac{a}{a+c}, \frac{BF}{c} = \frac{a}{a+b}$, so $CE = \frac{ab}{a+c}, BF = \frac{ac}{a+b}$, then $CE+BF = \frac{a(ab+ac+b^2+c^2)}{(a+b)(a+c)}$.
but $\angle A = 60^\circ$, so $b^2+c^2-2bc\cos60^\circ = a^2$, then $b^2+c^2 = a^2+bc$, so $CE+BF = \frac{a(a^2+ab+ac+bc)}{(a+b)(a+c)} = \frac{a(a+b)(a+c)}{(a+b)(a+c)} = a = BC$
