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From the figure,I could find that 1 and 3 are similar.Again how can I proceed?

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You are undoubtedly expected to use a similarity argument. So what is below is irrelevant to your needs.

We will use coordinate geometry. Choose $D$ as the origin, and let the axes be as usual. Let the side of the square be $1$ unit. The diagonal $AC$ has equation $x+y=1$. The line $DE$ has slope $1/2$, so it has equation $y=x/2$. The diagonal $AC$ and the line $DE$ therefore meet at $(2/3,1/3)$.

Now we know that Triangle $2$ has base $1$, height $1/3$, so area $1/6$. The areas of Triangles $1$ and $3$ are just as straightforward, and then we get the area of $4$ by subtraction.

Remark: The method of coordinates is a nice tool. Thank you, Fermat; thank you, Descartes.

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If triangles (or really any polygons) are similar and corresponding sides are in a ratio $a:b$, then their areas are in ratio $a^2:b^2$. From that, and the midpoint assumption, we can draw a conclusion about the ratios of the areas of $1$ and $3$. You also know that the area of $1,4$ together is equal to the area of $2,3$ together. Finally, what portion of the square's area is contained in $1,2$ together? That should suffice to get everything you need.

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Cameron’s given you the nice way to solve the problem. You can also solve it in an uglier but very straightforward way. Since you’re only interested in ratios, assume that the square is a unit square, so that $|AB|=1$, $|BE|=1/2$, and so on. enter image description here Using the similarity of $\triangle AHD$ and $\triangle CHE$, you can calculate $|JH|$ and $|HK|$ and from those the areas of $\triangle AHD$ and $\triangle CHE$. Knowing the value of $\angle HCG$, you can easily calculate $|HG|$ and then get the area of $\triangle DHC$. The area of the quadrilateral $ABEH$ is then just what’s left when the areas of the three triangular regions are subtracted from the area of the square, but you can get a double check easily enough: knowing $|HB|$, you also know $|FH|$, so you can calculate the areas of $\triangle AFH$ and the trapezoid $FBEH$.

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