Proof of the 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…

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.

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

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  • math/pythagoras/start.txt
  • Last modified: 2016-11-27 15:45 (5 years ago)
  • by Rik Blok