Look at it in another way

Construct a point $J$ on $\overrightarrow { HA }$ with distance $HF$ to the left of point $A$ . Then $BIHJ$ has rotational symmetry, so $\triangle HNG \cong \triangle BMA$ , so $[BMA] = 10$ .

Note that $\triangle BMC$ and $\triangle BMA$ have the same height if you take $\overline{MC}$ and $\overline{MA}$ to be the respective bases. Hence, $\frac{[BMA]}{[BMC]}=\frac{MA}{MC}$ .

Note that $\triangle BMC \sim \triangle HMA$ , so $\frac{MA}{MC}=\frac{MH}{MB}=\frac{BH}{BM}-1$ .

Note that $\triangle BMC \sim \triangle BHI$ , so $\frac{BH}{BM}=\frac{BI}{BC}=4$ .

Therefore, $\frac{[BMA]}{[BMC]}=4-1=3$ . But we know $[BMA]=10$ , so $\frac{10}{[BMC]}=3$ , so $[BMA]=10/3$ , so $m=10$ and $n=3$ , so $m+n=\boxed{13}$ .

By the intercept theorem it's easily seen that $(BGN)=9(BCM)$ and that $(BGH) = 12 (BCM)$ , so that $(NGH) = 3(BCM)$ and eventually $(BCM) = 10/3$ . $\square$

you can easily prove that $ABCD$ is a parallelogram.So $BC \parallel AD$ .repeating this for other triangles results that $A,D,F,H$ are collinear and $BI \parallel AH$ . So $ABGH$ is parallelogram.taking area of $\triangle ABC = x$ , area of $\triangle BGH = \frac{6x}{2}=3x$ .So area of $\triangle BHI = 4x$ . note that $AC \parallel HI \ \Rightarrow \triangle BMC \sim \triangle BHI \Rightarrow\left( \frac{BC}{BI}\right)^2=\frac{[BCM]}{[BIH]} =\frac{1}{16}$ . Now prove $\triangle BMC \cong \triangle FNH$ . By this we get the equation $\frac{x}{4}+10=x$ . using this we can find $x$ and then $\frac{x}{4}$ (area of $\triangle BMC$ )

First, is not hard to find out that $\bigtriangleup BCM \cong \bigtriangleup HFN$ . Now we set $\overline{BC}=b$ and $\overline{MK}=h$ where $MK$ is height of $\bigtriangleup BCM$ .

$[HFG]-[HFN]=10$ $\Rightarrow \frac{b\cdot 4h}{2} - \frac{bh}{2} = 10$ $\Rightarrow 4bh-bh=20$ $\Rightarrow \frac{bh}{2}=\frac{10}{3}$ $\Rightarrow n+m=10+3=13$