What the theorem actually says
Take a triangle with one right angle. Call the two short sides that meet at the right angle the legs, with lengths a and b, and call the side facing the right angle — always the longest — the hypotenuse, with length c. The Pythagorean theorem makes one sharp claim about these three lengths: a^2 + b^2 = c^2. The square built on the hypotenuse has exactly the same area as the two squares built on the legs put together.
Notice this is a statement about areas dressed up as a statement about lengths. The term a^2 is not just 'a multiplied by itself' in the abstract — it is the area of a literal square whose side is the leg a. So the theorem is a balance: the area of one big square on the left tray, the areas of two smaller squares on the right tray, and the two trays always level. That picture is worth holding onto, because most of the hundred proofs are really area-bookkeeping arguments in disguise.
The proof that grows out of similar triangles
Of all the proofs, the one that belongs in this rung uses the machinery you built in the last guide. Start with right triangle ABC, the right angle at C, and drop the altitude from C straight down to the hypotenuse AB, meeting it at a point H. That single line splits the big triangle into two smaller right triangles, triangle ACH and triangle CBH — and the surprise, which you proved as the altitude-on-hypotenuse relations, is that both small triangles are similar to the whole triangle and to each other.
Why are they similar? By AA similarity: triangle ACH shares angle A with the big triangle and has its own right angle at H, so two angles match and the triangles are similar. The same argument matches triangle CBH to the big one through the shared angle B. Now read off the proportions that similarity hands you. Each leg of the original triangle turns out to be the geometric mean between the whole hypotenuse and the piece of it nearest that leg.
Right triangle ABC, right angle at C, altitude CH to hypotenuse AB. Let c = |AB|, and split it as |AH| = p, |HB| = q, so p + q = c. Leg b = |AC| is the geometric mean of c and p : b^2 = c * p Leg a = |CB| is the geometric mean of c and q : a^2 = c * q Add the two: a^2 + b^2 = c*q + c*p = c*(p + q) = c * c = c^2.
Look at the last line. The two squared legs are each a slice of the hypotenuse times the whole hypotenuse, and when you add them the two slices p and q snap back together into the full length c. The whole theorem collapses into one honest line of algebra — no clever rearrangement of paper squares, just similarity and addition. This is why we placed it here: Pythagoras is not a separate marvel bolted onto geometry, it is what similar triangles say when one angle is a right angle.
The rearrangement proof, and why there are a hundred
Here is a second proof of a completely different flavour, the one most people meet first. Take four copies of the right triangle, legs a and b. Arrange them inside a big square of side (a + b) so that their hypotenuses face inward — they frame a tilted square of side c in the middle. Now slide the same four triangles into a different arrangement inside the same big square, and they instead leave two square holes, of sides a and b. Same big square, same four triangles removed: the leftover area must match. So c^2 = a^2 + b^2, read straight off the holes.
These two proofs feel nothing alike — one is pure proportion, the other is cut-and-slide area bookkeeping — yet both land on a^2 + b^2 = c^2. That is the real reason the theorem famously has hundreds of proofs (one 1940 collection gathered 370). A statement that ties together length, area, similarity, and the right angle can be reached from any of those directions. Even a young James Garfield, later a U.S. president, published his own using a trapezoid. The abundance is a sign of how central the result sits, not of any doubt about it.
Using it both ways: the converse and triples
A theorem and its converse are different claims, and Pythagoras has a useful one. The forward direction says: right angle implies a^2 + b^2 = c^2. The converse runs the other way: if three side lengths happen to satisfy a^2 + b^2 = c^2, then the triangle must be right-angled, with the right angle opposite the longest side. This is genuinely a separate fact needing its own proof, not the original read backwards — and it is what makes the theorem a test for right angles, not just a consequence of them.
The converse is the secret behind the carpenter's and builder's '3-4-5' trick. Because 3^2 + 4^2 = 9 + 16 = 25 = 5^2, a triangle with sides 3, 4, 5 is guaranteed right-angled — so you can lay out a perfect square corner on a building site with nothing but a knotted rope. Whole-number side sets like this are called Pythagorean triples; 5-12-13 and 8-15-17 are two more worth recognizing on sight, and any scaled copy like 6-8-10 works just as well.
- Identify the longest side; call its length c. The right angle, if any, must be opposite it.
- Compute a^2 + b^2 from the two shorter sides, and compute c^2 separately.
- If a^2 + b^2 = c^2 exactly, the triangle is right-angled (converse).
- If a^2 + b^2 > c^2, the angle opposite c is acute; if a^2 + b^2 < c^2, it is obtuse.
That last step is a quiet bonus: the same comparison not only detects right angles but classifies the triangle as acute or obtuse, because it is just the law of cosines telling you the sign of cos(C). So a^2 + b^2 versus c^2 is a single, cheap diagnostic for the largest angle's character.
Where the theorem carries you next
Pythagoras is one of those rare results whose reach keeps widening. The distance formula between two points in the plane is nothing but the theorem applied to the horizontal and vertical gaps: the straight-line distance is the hypotenuse of a right triangle whose legs are those two gaps. Stretch it into space and you get the distance in three dimensions; abstract it further and a^2 + b^2 = c^2 becomes the very definition of length in any number of dimensions. The little right triangle is hiding inside every measurement of how far apart two things are.
It also feeds straight into the trigonometry waiting one rung up. The Pythagorean identity sin^2 theta + cos^2 theta = 1 is Pythagoras applied to a right triangle inside a circle of radius 1, with the legs renamed sin theta and cos theta. The next guide stays right here, though, putting the theorem to work on two especially clean triangles — the 30-60-90 and the 45-45-90 — whose side ratios you will end up knowing by heart, plus how those ratios behave when you scale the whole figure up or down.