A) Let triangle $A B C$ be equilateral, i.e., $A B = A C = B C$. The midpoint of $A B$ is $M$. Then, by the SSS congruence criterion: $\triangle A M C \cong \triangle B M C$.

Thus, $\angle A C M = \angle B M C$, i.e., the median $CM$ is also an angle bisector. The same statement can be proven similarly for the other medians.

B) In triangle $A B C$, let $M$ be the midpoint of $A B$, and the median $C M$ is also an angle bisector. Then,

$$A M = M B \quad \text{(1)} \quad \text{and} \quad \angle A C M = \angle M C B$$

The parallels to $A C$ through $B$ and to $B C$ through $A$ intersect at a point $C^{\prime}$, and for this, $A C^{\prime} B C$ is a parallelogram. Due to (1) and because the diagonals of a parallelogram bisect each other, $C C^{\prime}$ passes through $M$.

Thus, $\angle M C B$ and $\angle A C^{\prime} M$ are alternate interior angles and therefore congruent; hence, and due to (2), $\angle A C M = \angle A C^{\prime} M$.

Therefore, triangle $A C^{\prime} C$ is isosceles with $A C = A C^{\prime}$. From this and $A C^{\prime} = C B$, it follows that $A C = B C$. (3)

Similarly, one can show: If, for example, the median through $B$ is also an angle bisector, then $A B = C B$. (4)

From (3) and (4), the equilateral nature of $\triangle A B C$ follows.