Sum of shaded areas: The area of a rectangle ABCD is 1. Assume E is on the diagonal AC...

[nanase]

Hello guys, I am trying to solve this geometrical question and is kinda stuck in the middle.

I tried labeling the sides of triangle and tried setting up equations in terms of pythagoras and areas

can you advise me on what I missed? thanks

 

[limiTS]

If the ratio of AE:EC is x:1, then similar triangles tell us that the ratio of FE:EG is also x:1. Likewise, the ratio of AF:GC is also x:1.

The diagonal AC, the vertical side BC=FG, and the horizontal side DC=AB can be divided into $\displaystyle\frac{x}{x+1}$ and $\displaystyle\frac{1}{x+1}$ proportional parts.

Let the rectangle's vertical side BC=FG be $l$. Since the rectangle's area is 1, then the horizontal side DC=AB must be $\displaystyle\frac{1}{l}$

Break the vertical and horizontal sides down into their proportional parts:

AF must be $\displaystyle\frac{x}{x+1}\cdot\frac{1}{l}$

GC must be $\displaystyle\frac{1}{x+1}\cdot\frac{1}{l}$

FE must be $\displaystyle\frac{x}{x+1}\cdot l$

EG must be $\displaystyle\frac{1}{x+1}\cdot l$

Now you have the perpendicular legs of both shaded triangles. Add up their areas. The $l$'s cancel, and you're left with an expression in $x$

 

[nanase]

If the ratio of AE:EC is x:1, then similar triangles tell us that the ratio of FE:EG is also x:1. Likewise, the ratio of AF:GC is also x:1.

The diagonal AC, the vertical side BC=FG, and the horizontal side DC=AB can be divided into $\displaystyle\frac{x}{x+1}$ and $\displaystyle\frac{1}{x+1}$ proportional parts.

Let the rectangle's vertical side BC=FG be $l$. Since the rectangle's area is 1, then the horizontal side DC=AB must be $\displaystyle\frac{1}{l}$

Break the vertical and horizontal sides down into their proportional parts:

AF must be $\displaystyle\frac{x}{x+1}\cdot\frac{1}{l}$

GC must be $\displaystyle\frac{1}{x+1}\cdot\frac{1}{l}$

FE must be $\displaystyle\frac{x}{x+1}\cdot l$

EG must be $\displaystyle\frac{1}{x+1}\cdot l$

Now you have the perpendicular legs of both shaded triangles. Add up their areas. The $l$'s cancel, and you're left with an expression in $x$

Ah yes I missed on using similarity concept, Thank you for giving me the sides, now I can find the areas of each triangle.
But I am curious in how you obtain those sides so I dig on the similarity ratios and got this $\frac{AE}{EC}=\frac{FE}{EG}=\frac{AF}{GC}$ . It seems you use different ratios to obtain the sides.
Thank you so much, after much scribbling and working backward from your expression I got the idea how the comparison of sides is made to obtain the expression for the base and height of each triangle.
It seems you are using
$\frac{AF}{AE}=\frac{DC}{AC}$

$\frac{GC}{EC}=\frac{DC}{AC}$

$\frac{FE}{AE}=\frac{CB}{AC}$

$\frac{EG}{AC}=\frac{DA}{AC}$

I fail to see and make ratios from those, but now I see it can be done. Thank you!