A line segment inside a square is perpendicular to another line segment that is also inside the square. Find the area in the diagram shown

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$ABCD$ is a square. $|AH|=2$ cm, $|EH|=6$ cm.

$FE||AB$

Find $A(ABEF)$.

There are only few known, so I tried to find some similarities by naming the angles in the right triangles, but I couldn't set up the ratios.

How can I solve this problem?

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Let $|AB|=:s$, $\>|BE|=:h$. Since $\triangle(AHD)\sim\triangle(EBA)$ we have $${2\over s}={h\over2+6}$$ and therefore $${\rm area}(ABEF)=hs=16\ .$$