I especially enjoyed the proof that similar right triangles have equal ratios of sides. Here is how I might expand it in a slightly.
Draw a rectangle and then add one of the diagonals.
This forms two right triangles which are congruent (by SSS or SAS).
Any rectangle is divided by its diagonal into equal areas above and below the line.
Next, introduce a point on the diagonal and draw two lines, one vertical and one horizontal. This forms more rectangles. Divide each of them along their diagonals (which lie along the original one).
All six triangles in the figure are similar, having all three angles equal. (Prove this using some combination of vertical and complementary angles and the alternate interior angles theorem).
As before, the two triangles shaded blue have equal area, as do the two shaded red.
The key step is to realize that since the whole area above the diagonal in the original rectangle is equal to that below, light blue is equal to dark blue, and light red is equal to dark red, the remaining areas are also equal.
tall skinny rect + light blue + light red = short fat rect + red + blue.
Coloring one of those sub-rectangles for clarity, we have shown (in other words) that white is equal to gray in the figure below.
The area in white is Ab and that in gray is aB. Equate them to obtain Ab = aB, and then divide, giving A/a = B/b.
Also, since A/a + 1 = (A + a)/a = B/b + 1 = (B + b)/b, either of the smaller triangles has the same ratios as the big ones that span the entire original rectangle.
Here is his figure:
From there, it is not too difficult to derive the Pythagorean theorem.
Except that first we need the converse theorem, which seems a bit tricky.
However, playing with the ratio above, we see that if Ab = aB, then not only A/a = B/b but also A/B = a/b.
Let A/a = k = B/b. Then we have that A = ka, and B = kb.
Place two right triangles with equal ratios of sides next to each other, and grow the small one by a factor of k by extending the base, preserving the acute angle at the base. Say we set the new length of the base to be ka.
Then, by the forward theorem we still have equal ratios, meaning that the height is equal to kb, which as we saw is equal to B. Therefore the top vertices superimpose at the same point.
Therefore, the two triangles are congruent, and equality of angles follows. Since we maintain equality of two of three angles in growing the triangle, we preserve all three.
Here is a proof without words for the Pythagorean theorem, growing triangles in the same fashion:
By a simple extension, the general result can be proved, all angles equal means equal ratios of sides, for any triangle.
k^2. (Relying on Euclid's famous I.47 which uses SAS).
Alternatively, here's a nice simple proof. In any right triangle, drop the altitude to the hypotenuse, h. This forms two more similar triangles.
h/x = y/h = b/a. But b/a is also the ratio of the hypotenuse when comparing the medium and small triangles, and the equality says that this ratio is the same as h/x the ratio of the short sides comparing the same triangles. This completes the proof.
