Problem 1455

1. Given Information and Initial Setup:
- We have an equilateral triangle $ABC$ with center $M$.
- Points $D$ and $E$ lie on $CA$ and $CB$ respectively such that $CD = CE$.
- $F$ is such that $DMFB$ is a parallelogram.
- We need to prove that $\triangle MEF$ is equilateral.

2. Properties of Equilateral Triangle and Parallelogram:
- Since $ABC$ is equilateral, $M$ is the centroid, and it divides each median in the ratio $2:1$.
- In parallelogram $DMFB$, opposite sides are equal and parallel, i.e., $DM = BF$ and $DF = MB$.

3. **Using the Given Condition $CD = CE$:**
- Since $CD = CE$, $\triangle CDE$ is isosceles with $D$ and $E$ equidistant from $C$.

4. Position of Points and Parallelogram Properties:
- Extend $BM$ to intersect $CA$ at $G$. Let $GM = 1$ and $MB = 2$.
- Since $DMFB$ is a parallelogram, $DF = MB = 2$.

5. Angle Considerations:
- Let $\angle GDM = x$. Then $MD = \frac{1}{\sin x}$.
- Since $DF \parallel BG$, $\angle GDF = 90^\circ$ and $\angle MDF = 90^\circ - x$.

6. **Using the Cosine Law in $\triangle DMF$:**
- By the cosine law:
$$MF = \sqrt{2^2 + \left(\frac{1}{\sin x}\right)^2 - 2 \cdot 2 \cdot \frac{1}{\sin x} \cdot \cos(90^\circ - x)}$$
- Simplifying the cosine term:
$$\cos(90^\circ - x) = \sin x$$
- Therefore:
$$MF = \sqrt{4 + \frac{1}{\sin^2 x} - 4 \cdot \frac{1}{\sin x} \cdot \sin x} = \sqrt{4 + \frac{1}{\sin^2 x} - 4} = \frac{1}{\sin x}$$
- Hence, $MD = MF = ME$.

7. Isosceles Triangle and Angle Sum:
- Since $MD = MF$, $\triangle DMF$ is isosceles, and $\angle DMF = 2x$.
- In quadrilateral $CDME$, the sum of angles is $360^\circ$. Thus:
$$\angle DME = 2x + 60^\circ$$

8. **Angle in $\triangle MEF$:**
- Since $\angle FME = 60^\circ$ and $FM = ME$, $\triangle MEF$ is equilateral.

$\blacksquare$