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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



math/pythagoras/start.txt · Last modified: 2016-11-27 15:45 (4 years ago) by Rik Blok

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