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As everyone knows, for a right triangle, the square on the hypotenuse is equal to the sum of the squares on the legs.

What happens if we replace the squares with some other geometrical figure, such as one arch of a cycloid? Does the Pythagorean relation still hold? That is, is the area under one arch of a cycloid that exactly fits the hypotenuse equal to the sum of the areas under the arches of the cycloids that exactly fit the legs?

It relies on the fact that area varies as square of scaling ratio.

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The theorem still holds. Suppose two geometric figures in the plane have the same shape. Then there is a function $f$ that maps every point in one of the two figures to a point in the other figure in such a way that there is some positive number $k$ such that for any two points $P,Q$ in the first figure, the distance between $f(P)$ and $f(Q)$ is $k$ times the distance between $P$ and $Q$. In other words, $f$ multiplies all distances by $k$.

A theorem of geometry says the area of the second figure --- the image of $f$ --- is $k^2$ times the area of the first figure --- the domain of $f$.

Thus the area of an arch of a cycloid is $k^2$ times the area of an arch of another cycloid if the length of the base of the first cycloid is $k$ times the length of the base of the second.

Say the lengths of the sides are $a,b,c$ and we have $a^2+b^2=c^2$. Let (capital) $A$ be the area of the cycloid whose base has length (lower-case) $a$. Then the cycloids whose bases have lengths $b$ and $c$ must have areas $(b^2/a^2)A$ and $(c^2/a^2)A$. So the areas of the cycloids are $$ A,\qquad \frac{b^2}{a^2} A, \qquad \frac{c^2}{a^2} A. $$ The theorem then entails that the sum of the first two of these is the third. Thus it works for arches of cycloids. And you can say the same if any other shape is used.

The aforementioned theorem of geometry doesn't get as much attention in classrooms as it deserves.

Here is a question I posted on using other shapes than squares to simplify the proof of the Pythagorean theorem. I was stunned by answer: Albert Einstein showed what may be the simplest of all proofs by using certain triangles instead of squares.

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    $\begingroup$ I would say, by the way, that it gets more attention than it used to. My son's textbook spends a non-trivial amount of time on it (not as rigorous theorem, but simply as a generalization). $\endgroup$ – Brian Tung Mar 3 '16 at 1:06
  • $\begingroup$ Is that a typo? 'c' should be 'k'? $\endgroup$ – user27325 Mar 3 '16 at 16:08
  • $\begingroup$ @EsperantoSpeaker1 : Gxuste kiel vi diras. Mi korektigis gxin. $\qquad$ $\endgroup$ – Michael Hardy Mar 3 '16 at 19:18
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There is absolutely no change. If similar figures are erected on the three sides, the area of the figure erected on the hypotenuse is the the sum of the areas of the two figures erected on the legs. For if the linear dimensions of a figure are scaled by a scaling factor $\rho$, then the area is scaled by $\rho^2$.

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Indeed it is, and it is a trivial matter to verify it. A full discussion is given here.

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Only the figures need be similar and closed. They can be closed semi-circles,ellipses, parabolic or cycloidal arches,sine curves or discontinuous boundary enclosures, even any three (Pythagorean relation obeying) scaled arbitrary polynomial similar figure plots spanning on such sides,.. and so on.

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