We first fold a square piece of paper in the middle, so that two congruent rectangles are created.
Then we fold $A$ onto midpoint $B$ of side $EC$ and mark points $D$, $F$, $G$ and $I$.
(a) Show that the triangles $\triangle IHG$, $\triangle BDC$ and $\triangle BEF$ are similar.
(b) Show that $|BD|:|BC|:|CD|=3:4:5$.
(c) Show that $|HF|=\frac{1}{3}\cdot |HE|$.
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At (a) we have that the triangles $\triangle BDC$ and $\triangle BEF$ are similar because:
The angles $\angle BEF$ and $\angle BGD$ are equal , they are both right angles.
It also holds that $|AG|=|BD|+|DG|$, $|EH|=|EF|+|FG|+|GH|=|EF|+|GI|+|GH|$ (since $|FG|=|GI|$, or not?), $|EG|=|EB|+|BG|=2|EB|=2|BG|$.
So we get that $\frac{|EB|}{|EF|}=\frac{|GB|}{|EF|}$. How can we continue?
I'm not sure if showing the sides being proportional is possible (at least I haven't put much effort to it yet. I might come back to it). The angle-angle(-angle) approach seems easier.
Also, I think you've typo'd. It should be $ \lvert AC\rvert = \lvert BD\rvert + \lvert DC\rvert $, etc.
Note that $\angle DBF$ is also a right angle, which, by symmetry, is the same as the right angle at A.
$\angle EBF = 90^\circ - \angle DBC$,
and hence $\angle EFB = 90^\circ - \angle EBF = \angle DBC$.
Together with the right angles at E and C, we have $\triangle BDC \sim \triangle FBE$.
As to why $\triangle IGH$ is also similar to the two triangles mentioned, think about the small triangle that went out of the square after folding the paper, say we call it $\triangle FGX$. By symmetry, $\triangle FGX \cong \triangle IGH$. Then it should be pretty straight-forward to show that $\triangle FGX \sim \triangle FBE$.