Date: 11/18/2002 at 06:19:58 From: Lina Subject: Symmedian point I found the symmedian point's definition as below: In the plane of any triangle ABC, let G be the centroid, and let La, Lb, and Lc be the bisectors of angle A, angle B, and angle C. Ga, Gb, and Gc are reflections of line AG about La, line BG about Lb, and line CG about Lc. The three lines Ga, Gb, Gc meet in the symmedian point. Can you prove it? Thanks.
Date: 11/18/2002 at 09:08:06 From: Doctor Floor Subject: Re: Symmedian point Hi, Lina, Thanks for your question. To prove that the three lines Ga, Gb, and Gc are concurrent, we use the trigonometric form of Ceva's theorem. See, for instance, from Alexander Bogomolny's site "Cut the Knot": Trigonometric form of Ceva's theorem http://www.cut-the-knot.org/triangle/TrigCeva.shtml This theorem says that if we have a triangle ABC, and on BC, AC, and AB respectively lie the points D, E, and F, then AD, BE, and CF are concurrent if and only if sin(<ABE)/sin(<CBE) * sin(<BCF)/sin(<ACF) * sin(<CAD)/sin(<BAD) = 1 This is rather, but not completely, similar to the "standard" Ceva's theorem, see from the Dr. Math archives: Ceva's Theorem http://mathforum.org/library/drmath/view/55095.html Now, in the above let D, E, and F be the midpoints of the sides, so that we know that AD, BE and CF concur in the centroid G. We then know that the identity indeed holds. If then we reflect AD about the bisector La of angle A, then this line meets BC at point D'. Note now that <CAD = <BAD' and <BAD = <CAD', because La divides angle A into two congruent angles. From this we see that sin(<CAD')/sin(<BAD') = sin(<BAD)/sin(<CAD) = 1/(sin(<BAD)/sin(<CAD)) In a similar way we find like expressions with E' and F', and we conclude that sin(<ABE')/sin(<CBE')*sin(<BCF')/sin(<ACF')*sin(<CAD')/sin(<BAD') = 1/(sin(<ABE)/sin(<CBE)*sin(<BCF)/sin(<ACF)*sin(<CAD)/sin(<BAD)) = 1/1 = 1 as desired. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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