After Michael Hardy, "Pythagorean Theorem Made Difficult", 1988

The Leg That Grows

Michael Hardy · 1988 · CalculusDifficultyPen & paper

If a right triangle has a fixed leg $b$ and a variable leg $a$ , its hypotenuse $c$ , treated as a function of $a$ , satisfies $c d c = a d a$ — and integrating that, with the right boundary condition, gives $c^{2} = a^{2} + b^{2}$ .

Take a right triangle with its right angle at $R$ . Hold the leg $b = R P$ fixed, and let the other leg, $a = R Q$ , be free to grow.
Let $a$ increase by a small amount $d a$ : $Q$ slides outward along the leg, from $Q_{0}$ to $Q$ , and the hypotenuse swings out to a new length, $c + d c$ .
Drop a perpendicular from the old point $Q_{0}$ onto the new hypotenuse $P Q$ , landing at $F$ . That gives a second, much smaller right triangle: $Q_{0} F Q$ .
$Q_{0} F Q$ shares its angle at $Q$ with the big triangle $R P Q$ , and both carry a right angle — so by AA, the sliver is similar to the whole triangle. That similarity gives $F Q / Q_{0} Q = R Q / P Q$ , i.e. $F Q / d a = a / c$ .
As $d a$ shrinks toward zero, $F$ slides along the hypotenuse to meet $Q$ , and the little segment $F Q$ becomes exactly $d c$ , the true increment in hypotenuse length. So $d c / d a = a / c$ , or written the other way, $c d c = a d a$ .
Integrate both sides as $a$ sweeps from $0$ up to its current value, accumulating every one of these infinitesimal slivers along the way: $\int c d c = \int a d a$ gives $\frac{c ^{2}}{2} = \frac{a ^{2}}{2} + C$ .
One boundary condition pins down $C$ : when $a = 0$ , $Q$ has slid all the way back to $R$ , the triangle has collapsed to nothing, and $c$ is just $b$ . Plugging that in: $C = b^{2} /2$ .
Substitute $C$ back in and multiply through by $2$ : $c^{2} = a^{2} + b^{2}$ , reached without ever drawing a square.

Therefore c² = a² + b². ∎

The similar-triangle step is the whole engine here, and it’s worth noticing it isn’t approximate — $FQ/Q_0Q$ equals $a/c$ exactly, for any $da$ at all, because $Q_0$ , $R$ , and $Q$ all sit on the same straight leg. What’s approximate is only the second half of that step, where $FQ$ is identified with $dc$ : that identification gets better and better as $da\to0$ , and is exact only in the limit. Everything before the integral is ordinary Euclidean similar-triangle reasoning; the calculus starts precisely at the word “integrate.”