Square

Let the square be $ABCD$ with side length $2a$ , the midpoints be $M$ and $N$ , and $AN$ cuts $DM$ at $P$ .

We note that $\triangle AMP$ , $\triangle APD$ and $\triangle AMD$ are similar 1-2- $\sqrt 5$ triangles. Therefore, the areas are directly proportional to the squares of their hypotenuse. Therefore, $[AMP]:[APD]:[AMD] = a^2: (2a)^2: (\sqrt 5 a)^2 = 1:4: 5$ . Since $[AMD] = \frac 12 (a)(2a) = a^2$ , then $[AMP] = \frac 15 a^2$ and $[APD] = \frac 45 a^2$ .

Now, we have:

$\begin{cases} A_{\color{#D61F06}red} = [MBNP] = [ABN] - [AMP] = \frac 12 (a)(2a) - \frac 15 a^2 = \frac 45 a^2 \\ A_{\color{#3D99F6}blue} = [DNP] = [ADN] - [APD] = \frac 12(2a)(2a) - \frac 45 a^2 = \frac 65 a^2 \end{cases}$

$\implies \boxed{A_{\color{#3D99F6}blue} > A_{\color{#D61F06}red}}$ .