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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
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

In a similar way we find like expressions with E' and F', and we
conclude that

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

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