Solution:

Let $AB = 2x$ and $AC = BC = y,$ so $BD = \frac{y}{4},$ $CD = \frac{3y}{4},$ and $\cos B = \frac{x}{y}$ (drop the altitude from $C$ to the midpoint of $\overline{AB}$). The law of cosines in triangle $ABD$ gives $$AD^2 = 4x^2 + \frac{y^2}{16} - 2 \cdot 2x \cdot \frac{y}{4} \cdot \frac{x}{y} = 3x^2 + \frac{y^2}{16}.$$

Both $CE$ and $BE$ are medians to $\overline{AD},$ in triangles $ACD$ and $ABD$ respectively. The median formula $4m^2 = 2b^2 + 2c^2 - a^2$ gives $$28 = 2y^2 + \frac{18y^2}{16} - AD^2 = \frac{49y^2}{16} - 3x^2, \qquad 36 = 8x^2 + \frac{2y^2}{16} - AD^2 = 5x^2 + \frac{y^2}{16}.$$ From the second equation $\frac{y^2}{16} = 36 - 5x^2;$ substituting into the first gives $49(36 - 5x^2) - 3x^2 = 28,$ so $248x^2 = 1736,$ $x^2 = 7,$ and then $y^2 = 16.$

The altitude from $C$ has length $\sqrt{y^2 - x^2} = 3,$ so the area is $\frac{1}{2} \cdot 2x \cdot 3 = 3\sqrt{7},$ and $m + n = 3 + 7 = 10.$