Square Division

Note that B and D are similar with ratio 2:1 [EH:MF=2:1], so their area is 4:1 [squared].

Since the edges of B and D are in ratio of 2:1, the diagonal [HJ:FJ] is divided into 2:1.

Therefore, B:C = 2:1.

From B:D=4:1 and B:C=2:1 we obtain C:D=2:1.

Since B+C is half the square, B+C= $\frac12$ , so B= $\frac13$ and C= $\frac16$ .

Since C:D=2:1, D= $\frac1{12}$ .

Therefore, A = $1-\frac13-\frac16-\frac1{12}$ = $\frac5{12}$ .

Therefore, A:B:C:D = $\frac5{12}:\frac13:\frac16:\frac1{12}$ = $5:4:2:1$ .

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Alternatively: (please enlarge this photo to view clearly)

Extend EM and HG to I.

Then, C is similar to (A+E) with a ratio 1:2, so the area ratio is 1:4 [squared].

And then?

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Note: the [square brackets] are annotations.

WLOG, Let the square have side length one and the lower left corner be the origin. Then the diagonal has equation $y=1-x$ and the second line has equation $y=\frac{x}{2}$ . These intersect where $1-x=\frac{x}{2} \implies x=\frac{2}{3}$ so they intersect at the point $\big(\frac{2}{3},\frac{1}{3}\big)$

Now:

$b=\frac{2}{3}\cdot\frac{1}{2}=\frac{1}{3}$

$c=\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}$

$d=\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{12}$

and $a$ is the remainder of the square with area $1$ meaning it must be $\frac{5}{12}$

Multiplying through by $12$ we have $a:b:c:d=5:4:2:1$ meaning the answer is $5+4+2+1=\boxed{12}$

The two lines trisect one another. Triangles C and D has the same height. So C=2D. Same with C and B. B=2C=4AD. A+D=B+C. So A=4D+2D-D=5D. A+B+C+D=5D+4D+2D+D= $~~~~\color{#D61F06}{12} * D.$