The Pythagorean Theorem

The statement, precisely

Take any right triangle — the right angle is the one that matters most, and the side opposite it is called the hypotenuse. The other two sides are the legs. Labeling:

• $a, b$ = the legs
• $c$ = the hypotenuse (always opposite the right angle)

Pythagorean Theorem. For any right triangle,

$a^2 + b^2 = c^2.$

That's the whole thing. But why is it true? What the actual hell do squared side lengths have to do with a right angle? The answer: squares of lengths are areas. The theorem says the area of the square on the hypotenuse equals the combined area of the squares on the two legs. Now it's geometrically obvious — or it will be, after the proof.

Proof 1: the rearrangement proof (four triangles in a big square)

This is one of the most beautiful proofs in mathematics, and it fits on a damn napkin. We're going to build the same area two different ways — no calculus, no tricks, just four triangles and some righteous algebra.

Setup. Draw a large square with side length $a + b$. Its area is $(a+b)^2$.

First way to tile it. Cut four congruent right triangles (legs $a, b$, hypotenuse $c$) from the corners, each with area $\tfrac{1}{2}ab$. What's left in the middle is a square with side $c$ (you should verify the angles work out — the corner angle from one triangle plus the right angle of the next is $a° + (90° - a°) = 90°$, so yes, the inner quadrilateral has all right angles and all sides equal to $c$). Its area is $c^2$.

So: $(a+b)^2 = 4 \cdot \tfrac{1}{2}ab + c^2$.

Second way to tile it. Rearrange the same four triangles into two rectangles (each $a \times b$) in two corners, leaving two uncovered squares: one of side $a$ and one of side $b$.

So: $(a+b)^2 = 4 \cdot \tfrac{1}{2}ab + a^2 + b^2$.

Conclude. The left sides are the same (same big square), so the right sides must be equal:

$4 \cdot \tfrac{1}{2}ab + c^2 = 4 \cdot \tfrac{1}{2}ab + a^2 + b^2.$

Subtract $2ab$ from both sides:

$c^2 = a^2 + b^2. \qquad \square$

The four triangles cancel and we're done. No algebra more powerful than subtraction was needed. That's it. That's the whole proof, and the Federation of Boring Textbook Authors still has the nerve to skip it. This is a proof about areas, and it works because area is additive.

Proof 2: the similar-triangles proof (altitude to the hypotenuse)

This is the proof that ties back to our last lesson — and if the first proof was beautiful, this one is fucking elegant. Draw right triangle $ABC$ with the right angle at $C$. Drop an altitude from $C$ to the hypotenuse $AB$, meeting it at $D$. This creates two smaller triangles: $\triangle ACD$ and $\triangle CBD$.

Claim: all three triangles are similar. Here's the AA argument:

• $\triangle ACD$ and $\triangle ABC$: share $\angle A$; both have a right angle (at $D$ and at $C$ respectively). AA similarity gives $\triangle ACD \sim \triangle ABC$.

• $\triangle CBD$ and $\triangle ABC$: share $\angle B$; both have a right angle (at $D$ and at $C$). AA gives $\triangle CBD \sim \triangle ABC$.

Now write the proportionality ratios. Let $AC = b$, $BC = a$, $AB = c$, $AD = p$, $DB = q$.

From $\triangle ACD \sim \triangle ABC$: $\frac{AC}{AB} = \frac{AD}{AC} \implies \frac{b}{c} = \frac{p}{b} \implies b^2 = cp.$

From $\triangle CBD \sim \triangle ABC$: $\frac{BC}{AB} = \frac{DB}{BC} \implies \frac{a}{c} = \frac{q}{a} \implies a^2 = cq.$

Add the two equations: $a^2 + b^2 = cq + cp = c(p + q) = c \cdot c = c^2. \qquad \square$

Because $p + q = AB = c$. Elegant. This proof shows you exactly where the Pythagorean theorem lives: in the self-similar structure of a right triangle.

The converse (and why it matters)

Converse. If $a^2 + b^2 = c^2$, then the triangle with those side lengths is a right triangle, with the right angle opposite the side of length $c$.

Why it matters: the original theorem lets you find side lengths when you know it's a right triangle. The converse lets you check whether a triangle is a right triangle when you only know side lengths. Used constantly in constructions, engineering, and the distance formula.

Proof sketch. Suppose $a^2 + b^2 = c^2$. Build a separate right triangle with legs $a$ and $b$; by the Pythagorean theorem, its hypotenuse is $\sqrt{a^2 + b^2} = \sqrt{c^2} = c$. So both triangles have the same three side lengths, making them congruent (SSS). The right angle in the constructed triangle corresponds to the angle opposite $c$ in the original. Done.

Pythagorean triples

A Pythagorean triple is a set of positive integers $(a, b, c)$ with $a^2 + b^2 = c^2$. The famous ones:

• $(3, 4, 5)$: $9 + 16 = 25$. The workhorse.
• $(5, 12, 13)$: $25 + 144 = 169$.
• $(8, 15, 17)$: $64 + 225 = 289$.

Any multiple of a triple is also a triple (scale by $k$: $(ka)^2 + (kb)^2 = k^2(a^2+b^2) = k^2c^2 = (kc)^2$). So $(6,8,10)$, $(9,12,15)$ etc. are the 3-4-5 family.

And $\sqrt{2}$ from Bedrock takes a bow here: the diagonal of a unit square has length $\sqrt{1^2 + 1^2} = \sqrt{2}$ — an irrational number, the first proof that not all lengths are rational (as the Pythagoreans discovered, to their considerable, shrieking horror — they reportedly threw the poor bastard who proved it overboard).

Payoff: the distance formula

This is the "later nodes" moment. In the Algebra Core lesson on the coordinate plane, the distance formula was stated as

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

Now you know why. The horizontal distance $|x_2 - x_1|$ and vertical distance $|y_2 - y_1|$ are the legs of a right triangle. The distance between the two points is the hypotenuse. The formula is the Pythagorean theorem, dressed in coordinates.

In the Matrices stratum, the length of a vector $\vec{v} = (v_1, v_2, \ldots, v_n)$ is $\|\vec{v}\| = \sqrt{v_1^2 + \cdots + v_n^2}$ — the Pythagorean theorem in $n$ dimensions. Same theorem, wearing a bigger costume.