A Triangle Vertex Bisection and Its Trio of New LengthsDate: 06/08/2012 at 11:28:37 From: Saurabh Subject: Length of angle bisector in a triangle In terms of its sides and trigonometric ratios, what will be the length of an angle bisector in any triangle having sides a, b, c and angles A, B, C? The most difficult part is finding the lengths in which the angle bisector divides the opposite side. Even if I assume them, and apply the law of cosines, simplification is very difficult. Date: 06/08/2012 at 13:08:30 From: Doctor Peterson Subject: Re: Length of angle bisector in a triangle Hi, Saurabh. See the following page, which gives a formula in terms of the side lengths only: http://mathworld.wolfram.com/AngleBisector.html I am not familiar with the formula given there for the length of the bisector; but the theorem on how it divides the opposite side is well known and easy to prove: http://mathworld.wolfram.com/AngleBisectorTheorem.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 06/09/2012 at 04:43:42 From: Saurabh Subject: Length of angle bisector in a triangle Thank you very much. The formula is useful. But what is the proof? Date: 06/09/2012 at 11:04:19 From: Doctor Peterson Subject: Re: Length of angle bisector in a triangle Hi, Saurabh. Are you asking for the proof of the formula I said I was unfamiliar with, or the proof I said was easy? I'll answer the latter, then see what I can come up with for the former. The Angle Bisector Theorem says The angle bisector of an angle in a triangle divides the opposite side in the same ratio as the sides adjacent to the angle. Before writing up the proof myself, I just searched our site and found it has already been done: The Angle Bisector and Equal Side Ratios http://mathforum.org/library/drmath/view/55014.html My preferred way to express this, using Doctor Santu's picture, is that the areas of two triangles with the same altitude are proportional to their bases. This ratio of ABD:CBD is, on one hand, equal to the ratio of bases AD:CD, with the red segment as their common altitude; and, on the other hand, to AB:BC, with the blue segments as their common altitude. Therefore, AD:CD = AB:BC a:b = x:y This is the form in which the theorem was stated above. Now, let's look at the other formula: The length t1 of the bisector A1T1 of angle A1 in the above triangle A1A2A3 is given by ... t1^2 = a2a3[1 - a1^2/(a2 + a3)^2], ... where ti = AiTi and ai = AjAk. To make things easier to write, let's set A1 = B A2 = A A3 = C T1 = D a1 = AC = b a2 = BC = a a3 = AB = c With this different labeling, we have B / \ / / \ c / \ a / / \ / t \ / / \ / x y \ A-----------D-----------C b This formula says The length t of the bisector BD of angle B in the above triangle BAC is given by t^2 = ac[1 - b^2/(a + c)^2] To prove this, we know the following: x + y = b x/y = c/a This is equivalent to ax - cy = 0 Solving for x and y, cx + cy = bc ax - cy = 0 ------------- (a + c)x = bc x = bc/(a + c) y = ab/(a + c) We can also apply the law of cosines to triangles ADB and BDC, as you suggested. If we call cos(ADB) "q," then cos(BDC) = -cos(ADB) = -q. So we have c^2 = x^2 + t^2 - 2xtq a^2 = y^2 + t^2 + 2ytq Multiplying the first by y and the second by x, then adding to eliminate q, we get xa^2 + yc^2 = xy^2 + x^2y + (x + y)t^2 Solving for t^2, t^2 = [xa^2 + yc^2 - (x + y)xy]/(x + y) t^2 = [xa^2 + yc^2 - bxy]/b Plugging in expressions for x and y, t^2 = [a^2bc/(a + c) + c^2ab/(a + c) - b(bc/(a + c))(ab/(a + c))]/b = [(a)abc/(a + c) + (c)abc/(a + c) - ab^3c/(a + c)^2]/b = [(a + c)abc/(a + c) - ab^3c/(a + c)^2]/b = (a + c)ac/(a + c) - ab^2c/(a + c)^2 = ac[1 - b^2/(a + c)^2] There's the formula! It sounds like your attempt was essentially the same, but less efficient and more error-prone. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 06/10/2012 at 04:08:50 From: Saurabh Subject: Thank you (Length of angle bisector in a triangle) Thank you very much!!! |
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