He didn't skip anything but he left the "obvious" details (for a mathematician) to the reader:
Let C be the area of the big triangle, A and B be the areas of the two small triangles. By construction we know that C = A + B. Moreover, a, b, c are the hypotenuses of the triangles A, B and C.
The area scaling quadratically with the similarity ratio means that
A = (a/c)² C, and B = (b/c)² C.
Now, plug this into A + B = C, cancel C, rearrange.
Mathematicians explain things the way I imagine musicians would if the ancient Greeks had insisted on making all musical instruments in a range audible only to dogs.
I'd be like, "How do I actually hear the difference between a major and minor sixth?" And the musician would be like, "Just play them into the cryptophone and note the difference in the way your dog raises its eyebrows."
The very few remaining musicians in this hellscape would be the ones who are unwittingly transposing everything to the human range in their sleep, then spending the day teaching from the Second Edition of the Principles of Harmonic Dog Whistling for all us schmucks.
Luckily we don't live in that musical universe. But mathwise, something like that seems to be the case.
Look, I think it's pretty hard for most of us to read long math arguments in plain text, so I wrote in the simplest language I could, leaving the simple details for the reader to fill in.
I will add that in the vast majority of mathematical literature, both in pedagogy and in research, the active participation of the reader is assumed: the reader is expected to verify the argument for themselves, and that often includes filling in the details of some simple arguments. That's exactly why math literature uses the plural first-person "we," because it's supposed to be as if the writer and reader are developing the argument together.
In contrast, listening to music can be purely passive (but doesn't need to be).
The thing is that in my head there is no algebraic argument: we go from (1) similarity ratios being A:B:C and (2) the first two areas adding up to the third area, straight to the conclusion of A^2 + B^2 = C^2. I think your point about a step being missing here is valid, but when I search my intuition, it's still not coming up as algebraic. I suspect this is the same for others like me who are inclined to think geometrically, but I'd like to hear their opinions.
Here's an attempt at filling in the geometric intuition with something more concrete. You know how it's common to visualize the theorem with squares on the three sides of the triangle and saying that the two small squares add up to the big one? And then everyone stares at it and says "huh?" because that fact is far from obvious from that diagram. Here's the thing though, we're free to choose different area units if we want. So just choose units where our triangle itself with a given hypotenuse H has area H^2 units. Then we can give the argument above without any extra factors and cancellations.
To fully justify the "choose any units," you do need to check that it's logically consistent, which you could say is more missing steps, but I think this idea is far more fundamental than the Pythagorean Theorem. Our use of squares to define the fundamental units of area really is a completely arbitrary choice. We call them "square units," which already biases us to think of area in a specific way, but there's absolutely no reason we can't use any other shape. Of course squares are convenient because you can stack them up neatly and count them, but that doesn't seem to be helpful at all in this context, so it's natural to choose something else.
> So just choose units where our triangle itself with a given hypotenuse H has area H^2 units.
This is not at all trivial. You're claiming you can choose units in such a way (reusing my notation from before) that simultaneously
A = a², B = b², C = c².
Intuitively, you can do that precisely because the triangles are similar and area is quadratic in the similarity ratio. But there is definitely some algebra behind that.
To be clear, I'm just claiming that we can choose a specific area unit, and the three equations you wrote are then obvious consequences of that. It's true, you do need to assume area scales as the square of length, but IMO that's a pretty fundamental fact, and I think that's intuitive for many others. But as always, YMMV.
The math is obvious enough, I agree. But the description of the approach feels like it's lacking something - specifically, something along the lines of "now write down the scaling equations and simplify the area summation." I feel like it's not at all clear they're switching to an algebraic argument there.