Blue or Red

Since no one provides a solution, I am submitting one. I was thinking of submitting a simple graphical solution but I couldn't find one.

Let the side length of the pentagon be $1$ and $DF=DG=a$ . Then the ratio of areas of the two triangle is given by:

$\frac {A_{\blue{\rm blue}}}{A_{\rm \red{red}}} = \frac {\frac 12\cdot (1-a) \sin 108^\circ}{\frac 12 \cdot a \cdot a \sin 108^\circ} = \frac {1-a}{a^2}$

To find $a$ , we note that $\triangle DFG$ and $\triangle DEH$ are similar. Therefore $\dfrac a1 = \dfrac {FG}{EH} = 1{2\cos 36^\circ} = \dfrac 1 \varphi$ , where $\varphi = \dfrac {1+\sqrt 5}2$ is the golden ratio . Then we have:

$\frac {A_{\blue{\rm blue}}}{A_{\rm \red{red}}} = \frac {1-a}{a^2} = \varphi^2\left(1-\frac 1\varphi \right) = \varphi^2 - \varphi = \varphi + 1 - \varphi = 1$ $\implies A_{\rm \blue{blue}} = A_{\rm \red{red}}$

Ans: Neither

Here is a geometric solution.

Draw the line through $D$ parallel to side $AB$ . Let this line intersect line $AE$ at $H$ . It is easy to compute that $\angle HAF = \angle DAF = 18^\circ$ and $\angle HDF = \angle ADF = 36^\circ$ , so $F$ is the incenter of triangle $ADH$ . This means $F$ is equidistant to lines $AH$ and $DH$ .

In triangles $AEF$ and $DFG$ , bases $AE$ and $FG$ have the same length. ( $AFGB$ is a rectangle, so $FG = AB = AE$ .) And from our observation above, the height from $F$ to $AE$ is equal to the height from $D$ to $FG$ . Therefore, triangles $AEF$ and $DFG$ have the same area.