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 $~BF + CE = BC$
 A: Let's give the intersection of the bisectors $BE$ and $CF$ a name: $S$. We draw the bisector of $S$ as follows (in blue):

It is clear, since $\angle BAC =60$, that $\angle ABC+\angle ACB=120$, and thus: $$\frac{\angle ABC+\angle ACB}{2}=\angle EBC+\angle FCB=\angle SBC+\angle SCB=60$$
Thus, $\angle BSC=120$ and from that follows easily that $\angle BSK=\angle CSK=60$, where $K$ is the intersection of the bisector of $\angle BSC$ and $BC$. Now note that $\angle BSK+\angle KSC+\angle CSE=180$, thus, we know $\angle CSE=60$. Now note that, in $\triangle SKC$ and $\triangle SEC$, we have $\angle KCS=\angle ECS$ and $\angle CSE=\angle CSK=60$, and so $\triangle SKC \sim \triangle SEC$, even, since they share edge $SC$, $\triangle SKC \cong \triangle SEC$. Thus follows $|KC|=|EC|$. Similarly, we can deduce $|FB|=|KB|$, and thus, 
$$|EC|+|FB|=|KC|+|KB|=|CB|$$
A: Here comes the simple solution:
Let $I$ be the intersection point of the angle bisectors $BE$ and $CF$. A direct angle chasing shows that $$\angle \, BIC = 120^{\circ} \,\, \text{ and } \,\, \angle \, BIF = \angle \, CIE = 60^{\circ}$$
Draw the lines $p$ through point $F$ orthogonal to the angle bisector $BE$ and let $G$ be the intersection point of $p$ with the edge $BC$. Since $BE$ is an angle bisector of angle $\angle \, B$
$$BF = BG$$ and so triangles $BFI$ and $BGI$ are congruent. Hence $\angle \, BIF = \angle \, BIG = 60^{\circ}$. For that reason 
$$\angle \, CIG = 180^{\circ} - \big( \angle \, BIF + \angle \, BIG \big) = 180^{\circ} - (60^{\circ} + 60^{\circ}) = 180^{\circ} - 120^{\circ} = 60^{\circ}$$
And thus $\angle \, CIE = 60^{\circ} = \angle \, CIG$. However $\angle \, ICE = \angle \, ICG$ because $CF$ is the angle bisector of $\angle \, C$. Therefore triangles $CIE$ and $CIG$ are congruent. Hence $CE = CG$. Finally
$$BC = BG + CG = BF + CE$$ 
A: 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 $CE+BF = \frac{a(a^2+ab+ac+bc)}{(a+b)(a+c)} = \frac{a(a+b)(a+c)}{(a+b)(a+c)} = a = BC$
A: The problem is fairly easy 
It can be easily shown that $\angle KBC+\angle KCB=60^\circ$
By Law Of Sines,
In $\triangle BFC$
$$\frac{BC}{\sin(\angle BFC)}=\frac{BF}{\sin(\angle FCB)}$$ which implies that 
$\frac{BC{\sin(\angle FCB)}}{\sin(\angle BFC)}={BF}$
Similarly,In $\triangle BEC$
$$\frac{BC{\sin(\angle EBC)}}{\sin(\angle BEC)}={EC}$$
Adding $BF$ and $EC$,
$$\frac{BC{\sin(\angle EBC)}}{\sin(\angle BEC)}+\frac{BC{\sin(\angle FCB)}}{\sin(\angle BFC)}$$
Taking BC common and keeping in mind that $\sin(\angle BFC)=\sin(\angle FBC+\angle FCB)$ and $\sin(\angle BEC)=\sin(\angle EBC+\angle ECB)$
$$\frac{{\sin(\angle EBC)}}{\sin(\angle BEC)}+\frac{{\sin(\angle FCB)}}{\sin(\angle BFC)}=1$$
from which we can conclude 
$BF+CE=BC$
