Let's assume $\angle A \geqslant \angle B \geqslant \angle C$.
First, we need to prove that $\triangle A B C$ must be an acute triangle, which only requires proving that $\angle A$ is an acute angle.
(1) If $\angle A$ is a right angle, then $A$ coincides with $E, F$, which contradicts $A E+A F = B C$.
(2) If $\angle A$ is an obtuse angle, then $E$ is on the extension of $C A$, and $F$ is on the extension of $B A$ (as shown in the figure). By the shortest distance from a point to a line, we have
$$B C > C E = A E + A C > A E + A F.$$

This contradicts $B C = A E + A F$.
Therefore, $\angle A$ is an acute angle, and $\triangle A B C$ is an acute triangle.
Next, we need to prove that $\triangle A B C$ must be an equilateral triangle, which only requires proving that $\angle A = 60^{\circ}$.

Let the three sides of $\triangle A B C$ be $a$, $b$, and $c$ (with $a$ being the side opposite $\angle A$).
By the property that the larger side is opposite the larger angle, we have
$$a \geqslant b \geqslant c.$$

Since $\triangle A B C$ is an acute triangle, we have
$$A E = c \cos A, \quad A F = b \cos A.$$

Substituting $A E + A F = B C = a$,
we get $\cos A = \frac{a}{b+c} \geqslant \frac{a}{2a} = \frac{1}{2}$.
Thus, $\angle A \leqslant 60^{\circ}$.
Therefore, $\angle B \leqslant 60^{\circ}$,
$$\angle C \leqslant 60^{\circ}.$$

Adding these, we get $180^{\circ} = \angle A + \angle B + \angle C \leqslant 180^{\circ}$.
Hence, (1), (2), and (3) must all be equalities. Therefore, $\triangle A B C$ is an equilateral triangle.