Triangle vs Quad

The octagon vertices may be obtained by two squares when one of them is rotated of 45°. In the figure the two squares are AELK and BFJD.

By simmetry ABC and BCG are congruent to BGE that is congruent o DKI that is congruent to DHI while DIJ has the same area of DKI then

Red Area = Area(ABC) + Area(BCG) = Area(BEFG)=AREA(DIJK)=Area(DIJ) + Area(DJK) = 2 Area(DIK) = Area(DHI) + Area(DIJ) = Yellow Area

The yellow triangle is composed of 2 quads and 1 right triangle (RT). If we look at the longer leg of yellow triangle, it's divided into two equal length by the 'diametrical' diagonal of the octagon. Therefore, the whole yellow triangle is a similar triangle to the smaller yellow RT with a scale of 2 (after considering that the shorter leg is on a shorter diagonal that's parallel to the diametrical diagonal earlier).

Now the red quad on the other hand, is composed of 2 RTs and 1 triangle. Since the 2 red RTs have 'diagonal chords' that 'contain' the original octagon's side each (meaning that they have the same angle) and they also shared the same longer leg, thus the two red RTs are congruent to each other (2 same angles + 1 same side). If you can see the comparison that the two red RTs together make up an isosceles triangle that's congruent to the white / uncolored isosceles triangle on the right side of the octagon by a mirroring effect (what is the name that other thing apart from translational & rotational movement?) along the axis that we get from connecting the top right vertex to the bottom left one. Afterwards, it should be easy to see that the two red RTs together is also congruent to the last red triangle from their same 2 octagonal sides containment (for a same apex angle) and the same 2 isosceles sides. Therefore, the whole red figure is twice the area of the isosceles triangle or four times the area of one red RT.

Playing jigsaw puzzle with the yellow parts, we can leave the bigger quad alone, move the smaller quad to the white / uncolored quad with 1 red edge & 1 yellow edge (mirror style) and the last yellow RT would have to be conveniently moved on top of the middle red RT, so we know that the RTs of different colours are congruent to each other. As both total are also 4 times the area of this 2 RTs (length scale of 2 equal area scale of 2² = 4), therefore the two figures have the same area.