Mike Staring's Calculus Based Proof

Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created October 21, 2025

Mike Staring published an interesting calculus based proof The Pythagorean Proposition: A Proof by Means of Calculus in 1996. We shall discuss Staring's proof on this page.

Let $ \bigtriangleup ABC $ be a triangle with $ \angle C = 90^{\circ} $, $ a = \overline{BC} $, $ x = \overline{AC} $ and $ f(x) = \overline{AB} $. Please see the dgaigram below. Here, we treat $ a $ as a constant and $ x $ as a variable so that the length of the hypotenuse $ \overline{AB} $ is a function $ f(x) $ of $ x $. Extending $ \overline{CA} $ to $ \overline{CD} $ so that $ \overline{AD} = \Delta x \geq 0 $. In this way, we have $ \overline{BD} = f(x+\Delta x) $. Let the circle with center at $ B $ and radius $ \overline{BD} $ meet $ \overleftrightarrow{BA} $ at $ F $. Let the circle with center $ B $ and radius $ \overline{BA} = f(x) $ meet $ \overleftrightarrow{BD} $ at $ E $. Hence, we have the following: $$ \begin{eqnarray*} \overline{BA} &=& \overline{BE} = f(x) \\ \overline{BD} &=& \overline{BF} = f(x+\Delta x) \\ \overline{AF} &=& \overline{ED} = f(x+\Delta x) - f(x) \end{eqnarray*} $$ Let the perpendicular foot from $ A $ to $ \overleftrightarrow{BD} $ be $ G $ and let the perpendicular foot from $ D $ to $ \overleftrightarrow{BF} $ be $ H $. Additionally, let $ \theta = \angle BAC $, let $ \alpha = \angle ADE $ and let $ \beta = \angle ABE $. Because $ \angle AEB = \frac{1}{2}\left(180^{\circ}-\beta \right)= 90^{\circ} - \frac{1}{2}\beta < 90^{\circ} $, $ G $ is between $ B $ and $ E$. Similarly, $ H $ is between $ A $ and $ F $.

The derivative of $ f(x) $ is \[ f^{\prime}(x) = \lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}. \]
Because $ H $ is between $ A $ and $ F $, we have \[ \frac{f(x+\Delta x) - f(x)}{\Delta x} = \frac{\overline{AF}}{\overline{AD}} \geq \frac{\overline{AH}}{\overline{AD}} = \cos(\theta) \] On the other hand, because $ G $ is between $ B $ and $ E $, we have \[ \frac{f(x+\Delta x) - f(x)}{\Delta x} = \frac{\overline{ED}}{\overline{AD}} \leq \frac{\overline{GD}}{\overline{AD}} = \cos(\alpha) \] Combining the above two gives \[ \cos(\theta)\leq\lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x) - f(x)}{\Delta x} = f^{\prime}(x) \leq \lim_{\Delta x \rightarrow 0}\cos(\alpha) = \lim_{\alpha\rightarrow \theta}\cos(\alpha) = \cos(\theta) \] Note that the last part above uses the fact of $ \cos() $ being a continuous function. From $ \bigtriangleup ABC $ we have $ \cos(\theta) = \frac{x}{f(x)} $. As a result, we have \[ f^{\prime}(x) = \frac{x}{f(x)} \mbox{\ \ \ \ \ or \ \ \ \ \ } f(x)\cdot f^{\prime}(x) = x \] For convenience, let $ y = f(x) $. We have \[ y\frac{dy}{dx} = x \mbox{\ \ \ \ \ or \ \ \ \ \ } (y) dy = (x) dx \] Integrating both sides yields \[ y^2= f^2(x) = x^2 + C, \] where $ C $ is a constant. Because $ f(0) = a $, we have $ C = a^2 $ and the Pythagorean Theorem $ x^2 + a^2 = f^2(x) $ follows directly.
For $ \Delta x < 0$, let $ x $ be $ \overline{CD}$. Then, $ \overline{AD} = x - \Delta x$ and $$ \begin{eqnarray*} f(x) &=& \overline{BD}=\overline{AF} \\ f(x- \Delta x) &=& \overline{BA}=\overline{BE} \\ f(x- \Delta x) - f(x) &=& \overline{AF}=\overline{ED} \end{eqnarray*} $$ The remaining is exactly the same and will not repeated here.
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