Steiner-Lehmus TheoremDate: 03/24/2000 at 09:16:40 From: Karri Chandra Sekhar Subject: Geometry 1. If in a triangle the lengths of two angle bisectors are equal, how can we prove that it is an isosceles triangle? (For this one, I tried and got an indirect solution. That is, I started at an assumption that resulted in a contradiction, so I have to change my assumption and try again to arrive at the correct solution. With this I am not satisfied. I need a direct solution. Can someone help me, please?) 2. There is a theorem that says that in a cyclic quadrilateral, the sum of opposite angles is 180 degrees. Is the converse of this theorem also true? That is, if in a quadrilateral, opposite angles add up to 180 degrees, can we prove that it is a cyclic quadrilateral? (For this one, it is enough to prove the converse of the following theorem: in a circle, if an arc forms 2x degrees at the center, it subtends an angle x on the circumference (i.e. at any point) above that arc. Kindly help me in solving these two problems. Thanking you, Chandra Sekhar Karri Date: 03/24/2000 at 16:24:00 From: Doctor Rob Subject: Re: Geometry Thanks for writing to Ask Dr. Math, Karri. You can prove the theorem directly by using the formula for the length of the angle bisectors. Let the lengths of the sides of the triangle be a, b, and c. Then let the bisectors of the angles opposite the sides of lengths a and b have lengths t_a and t_b. Then the formulas for t_a and t_b from the Dr. Math FAQ page of Triangle Formulas (scroll down to the scalene triangle) are: t_a = sqrt(b*c*[1 - a^2/(b+c)^2]) t_b = sqrt(c*a*[1 - b^2/(c+a)^2]) http://mathforum.org/dr.math/faq/formulas/faq.triangle.html Set these equal to each other. Square both sides, and multiply by (b+c)^2*(c+a)^2 to clear all fractions. (This factor can't be zero, so this is allowable.) Bring everything over onto one side of the equation, combine like terms, and factor the resulting polynomial. When I did this, I got c*(b-a)*(a+b+c)*(a^2*b+a*b^2+3*a*b*c+a*c^2+b*c^2+c^3) = 0 Now since all of a, b, and c are positive, the only way this equation could be true is if b - a = 0, that is, a = b. The converse of the theorem above, "In a circle if an arc forms 2x degrees at the center, it subtends an angle x on the circumference (i.e. at any point) above that arc," would say, "If the vertex of an angle of measure x lies on a circle, then the arc its subtends has measure 2*x at the center." Draw the angle and the circle. From the vertex of the angle, draw a diameter of the circle, and from the points of intersection of the sides of the angle and the circle, draw radii of the circle. Now the angle is the sum (or difference) of two angles a and b made by the sides of the angle and that diameter. Thus x = a + b, or x = a - b, or x = b - a, depending on the relative position of the diameter and the two sides of the angle. Each of the two triangles so constructed is isosceles, because two of their sides are radii of the circle, hence equal. Thus the base angles are equal, and so equal to a or b, respectively. Then the external angles at the center formed by the diameter and each of the radii are 2*a and 2*b, because they are the sum of the interior angles at the other two vertices of each triangle, which are both equal to a or b. Then the central angle subtending the arc is 2*(a+b) or 2*(a-b), or 2*(b-a), and so equal to 2*x. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 04/22/2000 at 05:38:50 From: chandra karri Subject: Re: Geometry Hello Dr. Math, First of all, let me thank you for your kind reply. As to the proofs of the problems I posed to you, in the case of first one, I need a proof based on pure geometry only (there should be no involvement of trigonometry.) You have supplied a proof that used expressions for the lengths of angle-bisectors. If possible, please send a direct proof based on existing theorems of geometry. The proof of second problem ... I could not follow it or visualize it. So if possible, please send the same proof with some illustrations. Then finally, I have a doubt about the definition of a point in geometry. How can we define a point in geometry? That's all for now, Thankingly, CS Karri. Date: 04/22/2000 at 11:53:42 From: Doctor Rob Subject: Re: Geometry The first problem is actually a famous theorem. It has a name: the Steiner-Lehmus Theorem. All the usual proofs are by contradiction, like your original one. The proof I gave using the formula for the lengths of the angle bisectors is direct. It is not trigonometric. It does use algebra, but not trigonometry. I'm sorry if it does not satisfy your stringent restrictions on the method of proof you will allow. The formula for the angle bisectors is a direct consequence of Stewart's Theorem, which says: Let d be the length of a line segment through C that divides the opposite side of length c into segments of length m and n adjacent to the sides of length a and b respectively. Then n*a^2 + m*b^2 = c*(d^2 + m*n). (See crudely drawn picture below.) C o / \\ / \ \ / \ \ a / d \ \ b / \ \ / m \ n \ o-------------o------o B c A Of course m + n = c. When the line segment of length d is the bisector of angle C, then n/b = m/a, which is a well-known theorem of geometry. Putting these together, you get c = m + n, = m*(1 + b/a) m = a*c/(a+b) n = b*c/(a+b) a^2*b*c/(a+b) + a*b^2*c/(a+b) = c*d^2 + a*b*c^3/(a+b)^2 a*b*c*(a+b)/(a+b) = c*(d^2+a*b*c^2/(a+b)^2) a*b = d^2 + a*b*c^2/(a+b)^2 a*b*(1-c^2/(a+b)^2) = d^2 That gives you the formula for the length of the angle bisector t_c = d of angle C as t_c = sqrt[a*b*(1-c^2/(a+b)^2)] Relabeling the sides gives the length of the angle bisectors of angles A and B as t_a = sqrt[b*c*(1-a^2/(b+c)^2)] t_b = sqrt[c*a*(1-b^2/(c+a)^2)] Thus, using just a little algebra, the Steiner-Lehmus Theorem is a consequence of Stewart's Theorem and other known theorems of plane geometry. There is also a fair amount of literature about this problem. Here are a few references turned up by a search: Court, College Geometry (1952), Barnes and Noble, p.72. H. S. M. Coxeter, Introduction to Geometry Second Edition, p. 420: He mentions that J. A. McBride has asserted that more than 60 proofs have been given. Coxeter and Geitzer, Geometry Revisited (1967), Random House, p. 15: Lehmus's proof of 1848. J. Fellman, Mathematics Student Journal (1962), p. 5. M. Gardner, Scientific American 204 (1961), pp. 166-168: with a follow up about 2 months later; the article in the book follows even this up in a footnote. (Book unknown). [I believe the book is titled New Mathematical Diversions: More Puzzles, Problems, Games, and Other Mathematical Diversions -Dr. TWE] G. Gilbert and D. MacDonnell, American Mathematics Monthly 70 (1963), pp. 79-80. Gilbert and MacDonnell, in the M.A.A. "Selected Papers on Geometry" (1979), pp. 127-8; reprinted from the American Mathematical Monthly 70 (1963), pp. 79-80. A. Henderson, Scripta Mathematica (1955), pp. 223-232, 309-312. (I have not seen this article, but the title is intriguing: The Lehmus-Steiner-Terquem problem in global survey.) M. Levin, Mathematics Magazine 47 (1974), pp. 87-89: he mentions that Rougevain was the first to publish a proof (1842). J. Malesevic, Mathematics Magazine (1970), pp. 101-102. J.A. McBride, Edinburgh Mathematical Notes 33 (1943), pp. 1-13. H.C. Overley, School Science & Mathematics (unknown date). A. Posamentier, Advanced Euclidean Geometry (1979), p. 72-75: 4 proofs. K. Seydel and C. Newman: Unknown, pp. 72-75 (after 1982; looks like an MAA journal, next article was a tribute to Beckenbach). For the second problem, here is a rough diagram of the case where the diameter lies between the sides of the angle: ,,--'--.. _,-' _,-o B / _,-' / \ . _,-' / . . _,-' / . A o---------o---------o D .`. O| . . `. | . \ `. | / `-._ `. | _,-' ``--o--'' C Here points A, B, C, and D lie on the circle. <CAB is the angle x you started with. O is the center of the circle. OA = OB = OC because they are radii of the circle. That means that <OAB = <ABO = (1/2)*<DOB, and <CAO = <OCA = (1/2)*<COD (since any external angle of a triangle is equal to the sum of the other two internal angles). Thus x = <CAB = <CAO + <OAB = (1/2)*(<COD+<DOB) = (1/2)*<COB 2*x = <COB The cases where the diameter AD lies outside the angle <CAB are much the same, but you have to subtract the angles <CAO and <OAB to get <CAB, and likewise you have to subtract the angles <COD and <DOB to get <COB. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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