Finding shaded triangle areas in a parallelogram There is the following parallelogram involving two shaded triangles.
If I found rightly, angles of $AMD$, $BMN$ and $CDM$ are $45$. But I can’t go further.
 
 A: $\angle AMD$, $\angle BMN$ and $\angle CDM$ are not necessarily $45^\circ$.

$[\triangle AMD]=\dfrac12\times\dfrac23\times[ABCD]$
$[\triangle BMN]=\dfrac12\times\dfrac13\times\dfrac13\times[ABCD]$
$[\triangle CND]=\dfrac12\times\dfrac23\times[ABCD]$
$[\triangle DMN]=\left(1-\dfrac13-\dfrac1{18}-\dfrac13\right)\times[ABCD]=\dfrac5{18}[ABCD]$
So, $\dfrac5{18}[ABCD]=\dfrac12\times6\times10$
$[ABCD]=108$
$\textrm{shaded area}=\left(\dfrac13+\dfrac1{18}\right)\times 108 = 42$
A: Since $MN||AC$, we obtain:
$$S_{AMCD}=\frac{DM\cdot AC}{2}=\frac{10\cdot18}{2}=90.$$
Let $S_{\Delta BMC}=s$.
Thus, since $$\frac{S_{\Delta AMC}}{S_{\Delta BMC}}=\frac{AM}{MC}=2$$ and $$S_{\Delta ADC}=S_{\Delta ABC}=S_{\Delta AMC}+S_{\Delta BMC}=2s+s=3s,$$ we obtain: $$S_{AMCD}=S_{\Delta AMC}+S_{\Delta ADC}=2s+3s=5s,$$ which gives $$s=18.$$
Now, $$\frac{S_{\Delta BMN}}{S_{\Delta BMC}}=\frac{BM}{BC}=\frac{1}{3},$$ which says $$S_{\Delta BMN}=6.$$
Also, $$\frac{S_{\Delta DMA}}{S_{\Delta DBA}}=\frac{MA}{BA}=\frac{2}{3}.$$
Thus, $$S_{\Delta DMA}=\frac{2}{3}S_{\Delta DBA}=\frac{2}{3}S_{\Delta ABC}=\frac{2}{3}\cdot3s=\frac{2}{3}\cdot3\cdot18=36$$ and since $$36+6=42,$$ we are done!
