Algebraic

Tip to Tail

Traditional · c. 1881 · AlgebraicDifficultyClassical

If $u$ and $v$ are perpendicular vectors, then $∣ u + v ∣^{2} = ∣ u ∣^{2} + ∣ v ∣^{2}$ — and a right triangle's three sides are exactly such a pair and their sum.

Let triangle $A B C$ have its right angle at $A$ , with $B$ and $C$ the other two corners.
Let $u$ be the vector running from $B$ to $A$ , and $v$ the vector running from $A$ to $C$ — two arrows, laid tip to tail.
The turn at $A$ is square, so $u$ and $v$ are perpendicular. That means their dot product is exactly zero: $u \cdot v = 0$ .
Follow $u$ , then $v$ , and you've walked straight from $B$ to $C$ . By the definition of vector addition, that path is $u + v$ — the hypotenuse, read as a single vector.
A vector's squared length is its dot product with itself. Expanding: $∣ u + v ∣^{2} = (u + v) \cdot (u + v) = u \cdot u + 2 (u \cdot v) + v \cdot v$ .
The middle term is the dot product from step three — and it's zero. It isn't cancelled against anything on the other side; it simply was never there.
What's left is $∣ u + v ∣^{2} = ∣ u ∣^{2} + ∣ v ∣^{2}$ . Since $∣ u ∣ = a$ , $∣ v ∣ = b$ , and $∣ u + v ∣ = c$ , that reads $c^{2} = a^{2} + b^{2}$ — reached without drawing a single square.

Therefore c² = a² + b². ∎

Nothing here is specific to triangles. The argument only ever uses that $\vec u\cdot\vec v=0$ for perpendicular vectors and that squared length means dotting a vector with itself — facts about vector algebra, not about right triangles in particular. The triangle is just the one picture where all three lengths happen to have names already: $a$ , $b$ , and $c$ .