Validity Michael Hardy's proof of Pythagoras Theorem using differentials

Question

A proof of the pythagorean theorem has been published by Mike Hardy during 1988 in Mathematical Intelligencer (Hardy, Michael, "Pythagoras Made Difficult". Mathematical Intelligencer, 10 (3), p. 31, 1988.). The proof can be found at https://en.wikipedia.org/wiki/Pythagorean_theorem under the section "Proof using Differentials".

In this proof, Hardy is using an approximation and the proof is based on the fact that two triangles are approximately similar due to very small differentials.

On the left image, the similar triangles are CDE and ABC. Based on that he derives into a differential equation that solving it produces the pythagorean theorem formula we all know.

My question is regarding the validity of the approximation. If the same way of though is used for any non-right triangle as seen in the right image, you can derive the pythagorean theorem formula for any triangle and thus make a wrong assumption as the pythagorean theorem does not stand for non-right triangles.

Thanks in advance.

I'm not sure if it's him or not. I stumbled upon the proof in wikipedia. The official link I was able to dig up was the publication in springer: Hardy, Michael, "Pythagoras Made Difficult". Mathematical Intelligencer, 10 (3), p. 31, 1988. — Κώστας Παπαστεργίου, Aug 14 '15 at 14:02
@AsafKaragila Really want to @ him but there is no way within this question : | — Vim, Aug 14 '15 at 14:14
By the way, I cannot find any reference of this paper on Wikipedia. — Asaf Karagila, Aug 14 '15 at 19:58

score 4 · Answer 1 · answered Aug 14 '15 at 14:14

If the two lengths marked $y$ are the same, the angles at $C, E$ are equal and "approximately" right angles - which is what gives the approximate similarity of the little triangle $CDE$ to the larger right-angled triangle $CBA$ in the first diagram (ignoring second order terms). There is no approximate similarity in the second diagram.

It would be interesting to see if the proof could be modified to prove the cosine formula by allowing for the deviation from similarity.

score 1 · Answer 2 · answered Aug 14 '15 at 14:18

The triangle $BEC$ approaches an isosceles triangle in the limit only if the angles at $E$ and $C$ are both right angles. So approximating the side lengths both to $y$ only makes sense the case of right-angles. This should have been made clear on the Wikipedia page.

Validity Michael Hardy's proof of Pythagoras Theorem using differentials

2 Answers2