Da Vinci’s Proof of the Pythagorean Theorem

Da Vinci’s proof of the Pythagorean Theorem is essentially the following picture:

Screenshot from 2015-09-07 22:26:05

The details of what you’re looking at are as follows:

Take a right triangle \Delta ABC (in black/gray) and augment each of its sides \overline{AB}, \overline{AC}, and \overline{BC} by adding the squares B_2=\square ABIH, A_2=\square AGFC, and C_2=\square BDEC, respectively, of side lengths |AB|, |AC|, and |BC|, respectively. Let a^2 = \text{Area}(A_2), b^2=\text{Area}(B_2), and c^2=\text{Area}(C_2). The goal, of course, is to show that a^2+b^2=c^2.

To do this, we let J=(x(D), y(E)) and we construct two auxiliary quadrilaterals \Omega_1=ABDJ and \Omega_2=FGHI (where here, x(P) and y(P) denote the x– and y-coordinates, respectively, of a point P in the plane).

Screenshot from 2015-09-07 22:43:20

The claim is that \omega_1=\text{Area}(\Omega_1) and \omega_2=\text{Area}(\Omega_2) are equal, which we deduce by proving that \Omega_1 is congruent to \Omega_2. Geometrically, this follows from the fact that quadrilateral \Omega_2 is nothing more than \Omega_1 reflected about the segment \overline{AJ} and rotated 90° about A.


Of course, this can also be argued rigorously using analytic/Euclidean-type arguments: See, e.g., here.

The end result is the same either way, of course: \Omega_1\cong\Omega_2 as claimed. Now, we can decompose the areas \omega_1 and \omega_2:

\begin{array}{rcl} \omega_1 & = & \frac{1}{2}\text{Area}(ABC)+\frac{1}{2}\text{Area}(BDEC)+\frac{1}{2}\text{Area}(DJE) \\[4pt] & = &\frac{1}{2}\text{Area}(ABC)+\frac{1}{2}c^2+\frac{1}{2}\text{Area}(ABC) \\[4pt] & = & \text{Area}(ABC)+\frac{1}{2}c^2\end{array},

\begin{array}{rcl} \omega_2 & = & \frac{1}{2}\text{Area}(AGFC)+\text{Area}(ABC)+\frac{1}{2}\text{Area}(ABIH) \\[4pt] & = &\frac{1}{2}b^2+\text{Area}(ABC)+\frac{1}{2}a^2\end{array};

In particular, \omega_1=\omega_2 yields the desired result: a^2+b^2=c^2. \square

Some people may not be happy with the filling in of \omega_i in the above blocked expressions: Visually, it’s not hard to believe that they decompose in such a way but Euclid would surely object. I myself am not Euclid, though I may go through and try to make some of these hand-wavey bits more Euclidean at some point.

Resources: For this, I made a pair of Geogebra worksheets, one simple and one more advanced. There exists at least one other such worksheet, too.