Proof of the Pythagorean Theorem
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
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
I found another proof, which Jim Loy (visit his Pythagorean page!) 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
Rik Blok