Looks like half the figure, right?

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Put $ABCD$ in a coordinate system as above

We have to find the intersection $O$ between the lines passing trhough $AN$ and $DM$

$\begin{cases} l_{AN}:y=2x\\ l_{DM}:y=-\dfrac{1}{2}x+1 \end{cases}$

$\Longrightarrow O=\left(\dfrac{2}{5},\dfrac{4}{5}\right)$

Using Gauss shoelace formula to calculate the area of $\triangle ODN$ :

$A_{ODN}=\dfrac{1}{2}\left|\dfrac{2}{5}+\dfrac{1}{2}-\left(\dfrac{2}{5}+\dfrac{4}{10}\right)\right|=\dfrac{1}{20}$

We can now calculate the area of $ABMO$ :

$\begin{aligned} A_{ABMO}&=\mathbin{\color{#D61F06}A_{ABCD}}- \mathbin{\color{#3D99F6}A_{ADN}}-\mathbin{\color{#20A900}A_{MCD}}+\mathbin{\color{#CEBB00} A_{ODN}}\\ &=\mathbin{\color{#D61F06}1}- \mathbin{\color{#3D99F6}\dfrac{1}{2}}-\mathbin{\color{#20A900}\dfrac{1}{2}}+\mathbin{\color{#CEBB00}\dfrac{1}{20}}=\dfrac{11}{20}=\dfrac{a}{b} \end{aligned}$

$\Longrightarrow a+b=11+20=\boxed{31}$

Let $P$ be the midpoint of $AD$ and $Q$ be a point on $BQ$ so that $MQ$ is perpendicular to $BP$ . Now note that $\angle BAP=\angle BQM=90$ degrees and $\angle ABP=\angle ABM-\angle QBM=90-\angle QBM=180-90-\angle QBM=180-\angle BQM-\angle QBM=\angle BMQ$ . As such, $\triangle ABP$ and $\triangle QMB$ are similar triangles.

Now let $BQ=x$ . If $AB=1$ , then $AP=BM=\frac{1}{2}$ . Then considering that $\triangle ABP$ and $\triangle QMB$ are similar, $\frac{BQ}{QM}=\frac{AP}{AB}=\frac{1}{2}$ . We have $\frac{x}{QM}=\frac{1}{2}$ therefore $QM=2x$ . Employing Pythagoras' Theorem:

$BQ^2+QM^2=BM^2$

$x^2+(2x)^2=(\frac{1}{2})^2$

$x^2+2^2x^2=\frac{1}{2^2}$

$(1+4)x^2=\frac{1}{4}$

$5x^2=\frac{1}{4}$

$x^2=\frac{1}{5\times4}$

$x^2=\frac{1}{20}$

Next $\frac{BP}{AB}=\frac{BM}{QM}$ or $\frac{BP}{1}=\frac{\frac{1}{2}}{2x}$ which means $BP=\frac{1}{4x}$ .

Letting $R$ be where $AN$ and $BP$ meet, we also see that $\triangle ARP$ and $\triangle MQB$ are congruent triangles so that $AR=QM=2x$ and $PR=BQ=x$ .

So area of $\triangle ARB=\frac{1}{2}\times AR \times BR=\frac{1}{2}\times 2x \times (\frac{1}{4x}-x)=x \times (\frac{1}{4x}-x)=\frac{1}{4}-x^2$ , area of $\triangle BQM=\frac{1}{2}\times BQ \times QM=\frac{1}{2}\times x \times 2x=x^2$ and area of rectangle $QROM=QM \times QR=2x(\frac{1}{4x}-2x)=\frac{1}{2}-4x^2$ .

Adding these three areas gives us $\frac{3}{4}-4x^2=\frac{3}{4}-\frac{4}{20}=\frac{3\times 5}{4\times 5}-\frac{4}{20}=\frac{15}{20}-\frac{4}{20}=\frac{15-4}{20}=\frac{11}{20}$

$a=11, b=20$ hence $a+b=11+20=\boxed{31}$