Triangle ConstructionDate: 09/09/2001 at 23:24:00 From: becca Subject: Geometry-triangles Given a triangle ABC and point D somewhere on the triangle (not a midpoint or vertex), I am having trouble constructing a line that bisects the area. I have been able to construct a line parallel to one side, but it must be at a constant ratio (1/SQRT 2). D can be anywhere. Using Sine ratios and the formula for the area of the triangle, I have discovered that, if D lies on BC and F is the point at which the line intersects AB, then BA/BF = 2BD/BC. (This assumes that D intersects AB, but a similar ratio can be derived if it intersects AC.) But I don't see how this can help me to construct the line. I've been working on this for days. I hope you can help! Many thanks. Date: 09/10/2001 at 04:05:20 From: Doctor Floor Subject: Re: Geometry-triangles Hi, Becca, Thanks for writing. Let me make a sketch to start with: C / \ / \ / E / \ A--D------B x y I have located D on side AB, and lengths AD = x, DB = y with x < y. In this case E must be on side BC. If x = y, then point E coincides with C, and when x > y, E should lie on AC. Now the region ADEC can be divided into two triangles by segment DC. Triangle ADC has area x --- * K x+y where K is the area of ABC. This can be seen by noticing that the heights of triangles ADC and DBC are equal, and their bases are x and y respectively. The areas of ADEC and DBE should be 1 (x+y)/2 - K = ------- 2 x+y and thus triangle DEC should have area (y-x)/2 ------- . x+y So the ratio of the areas of triangles DBE and DEC should be x + y : y - x. Then, again because these triangles have equal heights seen from bases DE and EC, this should give us that BE : EC = x + y : y - x. Now point A can be constructed in the following way: C \ E AE//XC \ \ X-------A----------------B y-x x+y Explanation: It is not too difficult to find point X and (since XB is 2y, point X is B reflected through D). Draw XC and let E be the intersection of BC and the line through A parallel to XC. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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