# Da Vinci’s Proof of the Pythagorean Theorem

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

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).

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.