Blue or Yellow

Let the side length of the larger equilateral triangle be $1$ . Then $AD = \frac 15$ and $DB = \frac 45$ . By cosine rule ,

$\begin{aligned} DE^2 & = \blue{EB}^2 + DB^2 - 2 \cdot \blue{EB} \cdot DB \cdot \cos B & \small \blue{\text{Note that }EB=AD = \frac 15} \\ DE^2 & = \frac 1{25} + \frac {16}{25} - 2 \cdot \frac 15 \cdot \frac 45 \cdot \cos 60^\circ \\ & = \frac {17}{25} - \frac 4{25} = \frac {13}{25} \end{aligned}$

Since the area of an equilateral triangle is directly proportional to the square of its side length, the area of the yellow equilateral triangle is $\frac {13}{25}$ that of the large equilateral triangle. This means the blue area is $\frac {12}{25}$ that of the large equilateral triangle. Hence yellow area is larger than the blue area.

Letting AD be 1 and DB be 4, the perpendicular distance of point F to the line AB is 2√3. We can easily get FD with √(FG² + GD²) = √( (2√3)² + (2 – 1)² ) = √13 . Compared with the original triangle with sides of 4 + 1 = 5, the ratio is ori : yellow = 5² : √13² = 25 : 13 and the blue : yellow should be 25 – 13 : 13 = 12 : 13