Pythagorean Theorem Proof by Brodie ExplainedDate: 03/15/2000 at 07:54:22 From: Anne Subject: A Pythagorean theorem proof by Dr. Scott Brodie Could you explain to me the proof of the Pythagorean theorem given by Dr. Scott Brodie using similar triangles? It would be helpful. Date: 03/15/2000 at 13:02:25 From: Doctor Peterson Subject: Re: A Pythagorean theorem proof by Dr. Scott Brodie Hi, Anne. You'll have to show me which proof you have in mind. I find his name associated with too many different web pages on the Pythagorean theorem to be sure which one you mean. Either tell me where to find it or send a copy. Also, tell me what part you have trouble with, so I don't have to re-explain the whole thing. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 03/15/2000 at 14:32:00 From: Anne Kolker Subject: Re: A Pythagorean theorem proof by Dr. Scott Brodie It is a proof using circles and similar triangles. On the page that I got it from, it was called Proof #22. It is not the one with the cosine rule. I can't remember the exact Web-address but it is a page with lots of different Pythagorean theorem proofs. I don't really understand any of it so explaining the whole thing would be great. I hope this helps. Date: 03/15/2000 at 17:04:48 From: Doctor Peterson Subject: Re: A Pythagorean theorem proof by Dr. Scott Brodie Hi, Anne. Okay, I guess you mean this proof, though it doesn't mention similar triangles: http://cut-the-knot.org/pythagoras/ Proof #22 Here is the second proof from the Dr. Scott Brodie's letter. We take as known a "power of the point" theorems: If a point is taken exterior to a circle, and from the point a segment is drawn tangent to the circle and another segment (a secant) is drawn which cuts the circle in two distinct points, then the square of the length of the tangent is equal to the product of the distance along the secant from the external point to the nearer point of intersection with the circle and the distance along the secant to the farther point of intersection with the circle. Let ABC be a right triangle, with the right angle at C. Draw the altitude from C to the hypotenuse; let P denote the foot of this altitude. Then since CPB is right, the point P lies on the circle with diameter BC; and since CPA is right, the point P lies on the circle with diameter AC. Therefore the intersection of the two circles on the legs BC, CA of the original right triangle coincides with P, and in particular, lies on AB. Denote by x and y the lengths of segments BP and PA, respectively, and, as usual let a, b, c denote the lengths of the sides of ABC opposite the angles A, B, C respectively. Then, x + y = c. Since angle C is right, BC is tangent to the circle with diameter CA, and the power theorem states that a2 = xc; similarly, AC is tangent to the circle with diameter BC, and b2 = yc. Adding, we find a2 + b2 = xc + yc = c2, Q.E.D. The first long paragraph states the following theorem, which I will fill out by labeling the points: Y+********** **** \ **** *** \ *** * \ * ** \ ** * \ * * \ * * \ * * + \ * * \ * * \* * X+\ ** ** \ * * \ *** *** \ **** **** \ ******+****---------------------+ Z W If a point W is taken exterior to a circle, and from the point W a segment WZ is drawn tangent to the circle at Z and another segment (a secant) WY is drawn which cuts the circle in two distinct points X and Y, then the square of the length of the tangent WZ is equal to the product of the distance WX along the secant from the external point W to the nearer point of intersection with the circle, X, and the distance WY along the secant to the farther point Y of intersection with the circle. WZ^2 = WX^2 * WY^2 He also assumes you know the theorem that a right angle can be inscribed in a semicircle (that is, if ABC is a right angle, then B lies on the circle whose diameter is AC): A *********** ***+ **** ** |\ ** ** | \ ** * | \ * * | \ * * | \ * * | + * * | \ * * | \ * * | \ * ** | \ ** ** | \ ** ***+-----------+*** B *********** C A final theorem he will use is that any line AB perpendicular to a radius OA of a circle, and passing through a point A of the circle, is tangent to the circle: \ \ \ \ *********** \ **** **** \ ** **\ ** ** * + A * / * * / * * + *\ * O * \ * * \ * * \ ** ** \ B ** ** **** **** *********** Now that we're ready, we can draw his figure: *********** **** **** *** *** oooo C ** ** ooooo ooo+o ** o **: o\ * o /* : o \ * o /* : o \ b * o a/ * : o \ * o / * : o \ * o / * : o \ * o / x **: o y \ * ooo+-----ooo+o----------------------------+* B oooo P ** c ** A *** *** **** **** *********** We start with a right triangle ABC, and draw the altitude CP from the vertex C to AB (perpendicular to AB), so angles ACB, BPC, and APC are all right angles. Now we construct circles with diameters AC and BC. The second theorem tells us that P is on both circles, since BPC and APC are right angles and can be inscribed in the semicircles. Now we label the lengths of segments BC (a), AC (b), AB (c), BP (x), and AP (y). Since P lies on AB, we know that x + y = c. We know that AC is perpendicular to BC; since BC is a diameter of one circle, the third theorem tells us that AC is tangent to that circle, while BC is tangent to the circle of which AC is a diameter. Now we can apply the first ("power of the point") theorem to circle AC and point B, with tangent BC and secant BA: BC^2 = BP * BA a^2 = xc and also to circle BC and point A, with tangent AC and secant AB: AC^2 = AP * AB b^2 = yc The last step is just to add these two equations: a^2 = xc b^2 = yc ------------------- a^2 + b^2 = xc + yc and then simplify the right side: a^2 + b^2 = (x + y)c = c^2 since x + y = c. That's what we wanted to prove. It's not an easy proof, but it brings in some of my favorite theorems! The same point P can be used in a much simpler proof that does use similar triangles. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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