Pythagorean Theorem - Euclid's ProofDate: 12/27/98 at 21:14:19 From: Gayle R. Subject: Pythagorean Theorem - Euclid's Proof Dear Dr. Math, The books I have read don't really explain Euclid's proof to the degree I need. I understand why they want to prove that the areas of the rectangles are equal to the squares of the legs of the right triangle, but I can't understand how they got to that conclusion. Thank you. Date: 12/28/98 at 11:52:16 From: Doctor Peterson Subject: Re: Pythagorean Theorem - Euclid's Proof Hi, Gayle. Euclid's proof on the Pythagorean Theorem is an interesting one. His proposition was: In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle. See David Joyce's Web presentation of Euclid's Elements: http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI47.html It sounds as if the part you are having trouble with is: Now the parallelogram BL is double the triangle ABD, for they have the same base BD and are in the same parallels BD and AL. And the square GB is double the triangle FBC, for they again have the same base FB and are in the same parallels FB and GC. Therefore the parallelogram BL also equals the square GB. The idea here is that we have a rectangle "BL" (BDLX) and a triangle ABD, and another rectangle "GB" (GFBA) and a triangle FBC. In each pair, we want to show that the rectangle has twice the area of the square; since we've proved that the two triangles are congruent (and therefore have the same area), the rectangles will therefore have the same area. It may be easier to follow by pulling those figures out from the whole figure: D B +---------+ | | \ | | \ +---------+-------+ L X A F B +------+ | |\ | | \ | | \ +------+---+ G A C The proof is the same for each pair. I'll work with the first. The area of the rectangle BDLX is twice the area of triangle ADB by Euclid's Proposition I.41 ("If a parallelogram has the same base with a triangle and is in the same parallels, then the parallelogram is double the triangle"); in modern terms, rectangle BDLX has twice the area of triangle BDL, which has the same area as ABD because they share the same base DB and the same altitude DL. If this still isn't enough detail, please write back and point out the part you don't understand, either from this copy of the proof or from the version you have. I'll be glad to help! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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