Proof of the Pythagoran theorem (1997)

Pythagorean theorem

In a right triangle the square of the hypotenuse is equal to the sum of the squares of the sides containing the right angle.

I’ve been puzzling over a proof for this for years, and it finally dawned on me. (Eureka!) It’s all in how you draw it…

Proof #1

https://docs.google.com/drawings/d/e/2PACX-1vRy7LCnp9BiaXX6iNwFM36CUb_8rxybJppPkCT8NK60fyrj0gRlry6KXHVWNKdIXYEq0-MOYrxQbpxX/pub?w=378&h=381 https://docs.google.com/drawings/d/e/2PACX-1vRy7LCnp9BiaXX6iNwFM36CUb_8rxybJppPkCT8NK60fyrj0gRlry6KXHVWNKdIXYEq0-MOYrxQbpxX/pub?w=378&h=381

Given the triangle formed by $a$, $b$ (choosing $b\geq a$) and $c$, we can construct a square with total area $c^2$. As shown, we can fit four triangles, each with area $a b/2$, into the large square, leaving an inner square with area $(b-a)^2$. Thus, the total area of the large square is

$$ \begin{array}{rl} c^2 & = 4 (a b/2) + (b-a)^2 \\ & = 2 a b + a^2 + b^2 - 2 a b \\ & = a^2 + b^2 . \end{array} $$

Hence, the Pythagorean theorem.

Proof #2

https://docs.google.com/drawings/d/e/2PACX-1vRIlckqYK8M0Qg22Ufgi2bv5LLNiSgroSoGceoIu9hapHHvhzgCSn-i5MdOKd2eIPzbm2tXXfAbdF0U/pub?w=302&h=132 https://docs.google.com/drawings/d/e/2PACX-1vRIlckqYK8M0Qg22Ufgi2bv5LLNiSgroSoGceoIu9hapHHvhzgCSn-i5MdOKd2eIPzbm2tXXfAbdF0U/pub?w=302&h=132

I found another proof, which Jim Loy told me is due to Legendre. It relies on recognizing that you can subdivide a triangle forming two sub-triangles similar to each other and the original. (I won’t prove this.) Then, from the figure above, and from the properties of similar triangles

$$ \frac{a}{e} = \frac{c}{a} \text{ thus } a^2 = c e $$

and

$$ \frac{b}{f} = \frac{c}{b} \text{ thus } b^2 = c f. $$

Adding the two results together gives

$$ \begin{array}{rl} a^2 + b^2 & = c e + c f \\ & = c (e+f) \\ & = c^2 . \end{array} $$

Hence, the Pythagorean theorem.

— Rik Blok, 1997