Find the area of the quadrangular region $ABNH $

For reference:

In the triangle $ABC$ is traced the height $BH$ . By the midpoint $M$ of $AC$ is traced to perpendicluar bisector that intercepts $BC$ in $N$. Calculate the area of the quadrangular region $ABNH$ if the triangle area $ABC= 18\ \mathrm{m^2}$ (Answer: $9\ \mathrm{m^2}$)

My progress:

$\triangle BCH \sim \triangle CNM \implies \dfrac{MN}{BH}=\dfrac{CM}{CH}$

$S_{ABNH}=18 -S_{ACN} = 18 - \dfrac{MN.CH}{2}$

$S_{BDN}=S_{DHM}\: (\text{by property})$

$S_{BDN}.S_{DHM}=S_{BDH}\cdot S_{DNM}\: (\text{by property})$

$\displaystyle \frac{S_{CMN}}{CN.CM}=\frac{18}{BC\cdot AC}=\frac{S_{BCH}}{CB\cdot CH}$

...?

Since $BM$ is a median, $[ABM]=9$. Also, $BH \parallel NM \Rightarrow [NBH]=[MBH]$.

$\therefore [ABNH]=[ABH]+[NBH]=[ABH]+[MBH]=[ABM]=9$.

Drop perp $NQ$ from $N$ to $BH$. Then, $S_{\triangle BNH} = \frac 12 BH \cdot MH ~$ (as $NQ = MH$)

$S_{ABNH} = S_{\triangle ABH} + S_{\triangle BNH} = \frac 12 BH \cdot (AH + HM)$
$= \frac 12 S_{\triangle ABC} = 9$