# De Gua and the Pythagoreans

The Pythagorean Theorem states that, given a right triangle, the areas of squares placed along the two legs will have the same area as a square placed on the hypotenuse. This is normally written as $$a^2 + b^2 = c^2$$, where $$a$$ and $$b$$ are the leg lengths and $$c$$ is the hypotenuse length.

De Gua’s Theorem (named for its first publication, by Jean Paul de Gua de Malves, in 1783, but known about much earlier) states that, given three right triangles with leg lengths such that we can form a tetrahedron, the sum of the squares of the areas of the three right triangles is equal to the square of the area of the base. Another way of describing this: Given a rectangular prism. Cut off a corner to form a tetrahedron. The sum of the squares of the areas of each face of the tetrahedron is equal to the square of the area of the base.

One proof of this involves Heron’s Formula (used to determine the area of a triangle given its side lengths), which is usually written as: $A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a + b + c}{2}$ but which can be written as $A = \frac{\sqrt{2(a^2b^2 + a^2c^2 + b^2c^2) – (a^4 + b^4 + c^2)}}{4}$

Label the legs of the right triangles as $$d = OA$$, $$e = OB$$, and $$f = OC$$, then label the hypotenuses as $$a = AB$$, $$b = BC$$, and $$c = CA$$. By the Pythagorean Theorem: $a^2 = d^2 + e^2 \\ b^2 = e^2 + f^2 \\ c^2 = f^2 + d^2$

Using Heron’s Formula, replacing these values, and simplifying gives the area of the base ($$\Delta ABC$$) as $A = \frac{\sqrt{d^2e^2 + e^2f^2 + f^2e^2}}{2}$ and its square as $A_{\Delta ABC}^2 = \frac{d^2e^2 + e^2f^2 + f^2e^2}{4}$

The area of each of the right triangle faces is: $A_{\Delta AOB} = \frac{de}{2} \\ A_{\Delta BOC} = \frac{ef}{2} \\ A_{\Delta COA} = \frac{fd}{2}$ The square of these, summed, is the square of the area of the base.

While the Pythagorean Theorem is biconditional (it is true of all and only right triangles), De Gua’s Theorem is not: It is possible to construct a tetrahedron where the square of the areas of three sides equals the square of the fourth. To see how, imagine that $$\Delta AOB$$ is equilateral. Varying the length of $$\overline{OC}$$ while keeping the other segments constant changes the areas of those two sides, while the base remains constant. For instance, set $$AB = BO = OA = 2$$ and $$CA = CB = 3$$. If $$OC = 1.6$$, the square of the base area is more than the sum of the squares of the side areas. If $$OC = 1.7$$, the square of the base area is less. That means that there’s some value $$1.6 \le x \le 1.7$$ where De Gua’s equation is satisfied but the sides are not right triangles.