Looking at the three trapezoids proof has inspired me to try my hand at this. The proof below is very unlikely to be new, but does not seem to be identical to any of the 80+ proofs listed already. It is actually a proof of the law of cosines, which of course includes the Pythagorean Theorem as a special case. It has the pleasant feature of treating all three sides of the triangle equivalently, the Law of Cosines only arising when any one side is singled out.

http://home.comcast.net/~stuartmanderson/images/Pythagorean.png

Let squares be erected on the sides of an arbitrary triangle ABC, and let altitudes be constructed from A, B, C to A', B', C' and extended to A", B", C" respectively. Assume that the triangle has no obtuse angles, so that lines A'A", B'B", and C'C" cut the three squares into rectangles. Since triangle AB'B and AC'C are similar, AB'/AB = AC'/AC, hence by cross multiplication, AB'·AC = AC'·AB, so that the two rectangles touching A have equal areas. This holds for each vertex.

Therefore, (AB)^2 = (AC)^2 - AC·B'C + (BC)^2 - BC·A'C. But B'C is by definition BC·cos(C) and A'C is by definition AC·cos(C), which gives the Law of Cosines immediately. The same relations hold true if signed areas are used for the case of an obtuse angle, hence the Law of Cosines follows in general. Of course, if one angle is right, then one pair of rectangles vanishes, and we immediately obtain the Pythagorean Theorem by decomposition of the largest square.

Stuart, thank you.

This is definitely not knew, but I never saw a historic attribution. We have a thread here

http://www.cut-the-knot.org/htdocs/dcforum/DCForumID11/90.shtml

Here's another one:

Inscribed "Star of David" proof of the Pythagorean Theorem:

http://stuartmanderson.home.comcast.net/images/diagram.png

Let a pair of mirror congruent right triangles ABC and DEF be inscribed in a circle with diameters AB and DE, so that DE is perpendicular to BC. (By symmetry, AB is perpendicular to EF.) Then DE bisects BC and its major arc, and likewise AB bisects EF and its major arc.

Then by the Broken Chord Theorem, EF cuts AB into parts (AB + AC)/2 and (AB - AC)/2, and by the Intersecting Chords Theorem, (AB + AC)/2 * (AB - AC)/2 = (EF/2) * (EF/2), which is equal to ( BC/2) * (BC/2). Therefore, (AB)^2 - (AC)^2 = (BC)^2, which is the Pythagorean Theorem.

--Stuart Anderson

Thank you. I think this is a distinctly worthy proof. It's going to be #88. Much simpler than #86: http://www.cut-the-knot.org/pythagoras/BrokenChordPythagoras.shtml

It is rather humbling to know that the three trapezoids "inspired" this. I have never seen anything like this, and it is quite elegant! Congratulations!

Well, I'm very happy that you both liked it! Upon reviewing the list of Pythagorean proofs, mine seems to be very similar to #79, but using the Broken Chord theorem instead of the dashed bisector line GI to establish the basic relation. I set out only to use the Broken Chord and Intersecting Chord theorems together; the nicely symmetrical diagram was just a bonus.

--Stuart Anderson

>... mine seems to be very
>similar to #79, but using the Broken Chord theorem instead
>of the dashed bisector line GI to establish the basic
>relation.

Thank you for the remark. I forgot about #79. Had to add a remark at the end of #88.

Just noticed that the page says "Start of David" instead of "Star of David" ...

Also, in perusing the list of proofs, I see that #23 refers to #7 as using Heron's formula, but this does not seem to be the case. Is there a typo in the number reference perhaps?

--Stuart Anderson

Thank you for both.

Heron's formula is used in the applet that illustrates the proof, not the proof itself. I hope that the present wording makes this clear.