Find the area of MBNO region below?

Question

For reference: Given a right triangle ABC straight in B. From the foot of the height BH trace the perpendiculars to AB and BC at the points M and N respectively. Calculate the area of MBNO region; if : AB = 7 and BC = 8 (O : point AC medium).(Answer:$14$)

My progress:

$\triangle ABC: = \frac{7.8}{2}=28\\ \triangle ABC: AC = \sqrt{8^2+7^2}=\sqrt{113}\implies\\ CO = AO = \frac{\sqrt{113}}{2}\\ \triangle CAB \sim \triangle CHN: \frac{\sqrt{113}}{AH}=\frac{8}{8-y}=\frac{7}{x}\\ OD \perp BC\implies OD = 3,5, DC =4$

...???

Answer 1

$\triangle AHB \sim \triangle ABC$. So, $\frac{AH}{BH} = \frac{7}{8}$

$\triangle AMH \sim \triangle BNH$, $\frac{AM}{BN} = \frac{AH}{BH} = \frac{7}{8}$

$S_{\triangle OAM} = \frac 12 \cdot OL \cdot AM = 2 AM = \frac{7}{4} \cdot BN$

$S_{\triangle OBN} = \frac 12 \cdot BN \cdot OD = \frac{7}{4} BN$

So, $S_{\triangle OBN} = S_{\triangle OAM}$

$S_{OMBN} = S_{\triangle OBM} + S_{\triangle OBN} = S_{\triangle OAB} = \frac{1}{2} \cdot 4 \cdot 7 = 14$

Answer 2

HINT.-For convenience, we translate the given figure to the figure below and calculate the coordinates of the points involved. This gives $$O=(0,0),A=(5.315,0),B=(-0.706,5.268),C=(-5.315,0)\\M=(1.905,2.948),N=(-3.317,2.284)$$ The required area is given by $$Area OMBN= area ABC-Area OMA-Area ONC$$ and this is easily calculate using vectors or the formula for area of a triangle in function of the coordinates of its vertex.

$$Area ABC=\frac{AB X AC}{2}=\frac12\begin{vmatrix}x_A&y_A&1\\x_B&y_B&1\\x_C&y_C&1\end{vmatrix}$$