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CTK Exchange
mr_homm
Member since May-22-05
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Sep-06-10, 06:52 AM (EST) |
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"Yet another Pythagorean Theorem proof"
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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. 
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. |
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mr_homm
Member since May-22-05
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Sep-17-10, 09:55 PM (EST) |
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2. "RE: Yet another Pythagorean Theorem proof"
In response to message #1
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Here's another one: Inscribed "Star of David" proof of the Pythagorean Theorem: 
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 |
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