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Garfield's Proof - June 10-14, 1996 Next Problem || Previous Problem || Contents || All Problems || Search POWs A while back the bonus question for the Problem of the Week was "Which U.S. President came up with a proof of the Pythagorean Theorem?" The answer, as many of you knew from the poster hanging in your classrooms, was James Garfield. (He actually came up with the proof before he became President.) This week, I want you to explain how the proof works. Here's the setup: Construct a right triangle. Label it ABC as shown. Using point B as a center, rotate side c and point A by 90 degrees (to get E). Connect points A and E and construct a line through point E parallel to side b. Extend side CB and label the point of intersection with the line through E as D. From all of this, you have to prove that a^2 + b^2 = c^2. (Hint: figure out all the areas possible using a, b, and c.) Good luck! Ultra-extra: Another U.S. president considered the study of Euclidean geometry to be very important to the development of the mind and critical thinking skills. Do you know who he was?) Proofs from Tim Ruttan, Amos Blackman, Del Rasouli Bin Abdul Rashid - who says that the answer to the Ultra Extra question is Woodrow Wilson (28th President of the USA), Ken Duisenberg, and Brian Gordon, who remembered already having sent us his proof in March. Tim Ruttan Grade: 11 School: Lakeridge High School Let us find the areas of triangles ABC, ABE, and BDE. Area(ABC) = .5*a*b Area(ABE) = .5*c*c, since the measure of angle ABE=90, and BE=c, since AB is rotated to form BE Area(BDE)= .5*a*b, since triangle BDE is congruent to triangle ABC by ASA - Let angle(CBA)=x; thus, by the Triangle Sum Theorem, angle(CAB)=90-x. - Angles CBA, ABE, and DBE are supplementary angles; thus they sum to 180. Since m(ABE)=90, CBA and DBE must add to 90; thus m(DBE)=90-x. - m(BDE)=90 by the perpendicular to parallels theorem, DBE and BED must add to 90. - Since m(DBE)= 90-x, m(BED)= x. - Thus, since m(BED)=m(CBA) and m(DBE)=m(CAB), and BE=CA, the triangles are congruent by the Angle Side Angle (ASA) theorem. Add the three area together: .5*a*b + .5*a*b + .5*c*c. This must equal the area of the trapezoid, which equals .5*(a+b)*(a+b). - DE=b by the congruent parts of congruent figures (CPCF) Theorem; DB=a; thus, CD=a + b. Thus, a*b + .5*c^2 = .5*(a^2+2*a*b+b^2) Multiply both sides by 2: 2*a*b + c^2 = a^2 + 2*a*b + b^2 Subtract 2*a*b from both sides: c^2 = a^2 + b^2 Pythagorean Theorem Proved. Amos Blackman Organization: Lakeridge High School Okay, here we go. Start with the original triangle, and since you rotate side c, BE = c, < (angle) ACB = 90, so since DE and AC are parallel, and DC is perpendicular so /\(triangle)ABC is a right triangle | /* m | c/ * so 2(.5ab)= area of /\BDE and /\ABC | / =ab. Area of trapezoid ACDE = | / * .5(a+b)(a+b), so area of /\ABE = | / .5a^2+ab+.5b^2 - ab = .5a^2+.5b^2. b| / * By the standard formula for the area | / of a triangle the area of /\ABE = | / * .5(c)(c) = .5c^2. So |/ .5a^2 + .5b^2 = .5c^2. Multiply it B o_ * by 2, and you get a^2 + b^2 = c^2 . | * _ | * _ c * Thanks, a | * _ Amos Blackman | * _ * C b A Del Rasouli Bin Abdul Rashid (18 years old) 95S11/S3A Tampines Junior College, Singapore Triangles ABC and BDE are congruent, because AB=BE=c, and angles ACB and BDE are right angles. Area of triangle ABC = Area of triangle BDE=1/2 ab Area of triangle ABE = 1/2 c^2 (Note: AB=BE=c) Area of trapezium AEDC = Sum of total areas of the 3 triangles 1/2 (a+b)(a+b) = 1/2 ab + 1/2 ab + 1/2 c^2 1/2 (a^2 + 2ab + b^2) = ab + 1/2 c^2 Multiplying by 2, we have, a^2 + 2ab + b^2 = 2ab + c^2 Therefore, a^2 + b^2 = c^2 (QED) Ken Duisenberg Grade: Hewlett Packard Engineer now. School: Stanford University '93, MSEE Solution: By rotating 90 degrees around B, triangle BED is formed identical to the original triangle ABC. That is: DE = BC = a, and DB = AC = b, and BE = AB = c. The area of the trapezoid ACDE is: height*(base1+base2)/2 = (a+b)*(a+b)/2 = (a^2 + b^2)/2 + ab The area of ACDE is also equal to the sum of the areas of ABC, BED and ABE: ABC: ab/2 BDE: ab/2 ABE: c*c/2 Setting these equal: (a^2 + b^2)/2 + ab = ab/2 + ab/2 + (c^2)/2 Canceling ab from both sides and multiplying by 2: a^2 + b^2 = c^2 Done. [Privacy Policy] [Terms of Use] Home || The Math Library || Quick Reference || Search || Help  © 1994-2008 Drexel University. All rights reserved. http://mathforum.org/ The Math Forum is a research and educational enterprise of the Drexel School of Education.