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Symmedian Point

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 

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 

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 

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 

 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 
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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