A geometry problem by Hemamalinivenkatasesha Nanduri

Construct $OG || AB.$
Let $CN=NB=BM=MA=a ,$ $OG=k$ and $NG=x.$
Since $\triangle ONG\sim \triangle ANB,$ using similarity properties,we get
$\dfrac{NG}{NB}=\dfrac{OG}{AB}\Rightarrow\dfrac{x}{a}=\dfrac{k}{2a}\Rightarrow k=2x.$
Since $\triangle OCG\sim \triangle MCB,$ using similarity properties,we get
$\dfrac{CG}{CB}=\dfrac{OG}{MB}\Rightarrow\dfrac{(a+x)}{2a}=\dfrac{k}{a}\Rightarrow a+x=2k\Rightarrow x=\dfrac{a}{3}$ and $k=\dfrac{2a}{3}.$
The area of $\triangle$ $OCN=\dfrac{1}{2}\times OG\times CN=\dfrac{a^2}{3}.$
The area of $\triangle$ $ANB=\dfrac{1}{2}\times AB\times NB=a^2.$
The area of the square $ABCD={(2a)}^2=4a^2.$
The area of quadrilateral $AOCD=$ Area of square $-$ Area of $\triangle OCN-$ Area of $\triangle ANB=\dfrac{8a^2}{3}.$
Therefore the ratio of the Area of quadrilateral $AOCD$ to the Area of the $ABCD=\dfrac{8a^2}{3}:4a^2=\dfrac{2}{3}=\boxed{0.666}.$

The red lines divide diagonal BD into three equal segments. (We can see this from the fact that BQ=QR because BN=NC, but RQ=RD because AP=PD.) So all six colored triangles in the image above are equal in size, as they have the same base along the diagonal and the same height relative to that diagonal.

Since the quadrilateral in question is made of 4 of those triangles out of 6, it takes up $\frac{2}{3}$ of the area of the square.