Editing Symmedian

{{Short description|Reflection of a triangle vertex's median over its angle bisector}}
[[Image:Lemoine punkt.svg|thumb|upright=1.25|
{{legend-line|solid grey|[[Median (geometry)|Median]]s (concur at the [[centroid]] {{mvar|G}})}}
{{legend-line|dashed grey|Angle bisectors (concur at the [[incenter]] {{mvar|I}})}}
{{legend-line|solid red|Symmedians (concur at the [[symmedian point]] {{mvar|L}})}}]]

In [[geometry]], '''symmedians''' are three particular [[straight line|lines]] associated with every [[triangle]]. They are constructed by taking a [[Median (geometry)|median]] of the triangle (a line connecting a [[Vertex (geometry)|vertex]] with the [[midpoint]] of the opposite side), and [[reflection (mathematics)|reflecting]] the line over the corresponding [[angle bisector]] (the line through the same vertex that divides the angle there in half). The angle formed by the '''symmedian''' and the angle bisector has the same measure as the angle between the median and the angle bisector, but it is on the other side of the angle bisector.

The three symmedians meet at a [[triangle center]] called the [[Lemoine point]]. Ross Honsberger has called its existence "one of the crown jewels of modern geometry".<ref name="h">{{citation|first=Ross|last=Honsberger|authorlink=Ross Honsberger|contribution=Chapter 7: The Symmedian Point|title=Episodes in Nineteenth and Twentieth Century Euclidean Geometry|publisher=[[Mathematical Association of America]]|location=Washington, D.C.|year=1995}}.</ref>

==Isogonality==
Many times in geometry, if we take three special lines through the vertices of a triangle, or ''[[cevian]]s'', then their reflections about the corresponding angle bisectors, called ''isogonal lines'', will also have interesting properties. For instance, if three cevians of a triangle intersect at a point {{mvar|P}}, then their isogonal lines also intersect at a point, called the [[isogonal conjugate]] of {{mvar|P}}.

The symmedians illustrate this fact. 
* In the diagram, the medians (in black) intersect at the [[centroid]] {{mvar|G}}.
* Because the symmedians (in red) are isogonal to the medians, the symmedians also intersect at a single point, {{mvar|L}}. 
This point is called the triangle's '''symmedian point''', or alternatively the '''Lemoine point''' or '''Grebe point'''.

The dotted lines are the angle bisectors; the symmedians and medians are symmetric about the angle bisectors (hence the name "symmedian.")

== Construction of the symmedian ==
[[Image:Symmedian_Construction.png|thumb|{{mvar|{{overline|AD}}}} is the symmedian through vertex {{mvar|A}} of {{math|△''ABC''}}.|alt=|275x275px]]
Let {{math|△''ABC''}} be a triangle. Construct a point {{mvar|D}} by intersecting the [[tangent]]s from {{mvar|B}} and {{mvar|C}} to the [[circumcircle]]. Then {{mvar|AD}} is the symmedian of {{math|△''ABC''}}.<ref>{{cite book |last1=Yufei |first1=Zhao |title=Three Lemmas in Geometry |date=2010 |page=5 |url=http://yufeizhao.com/olympiad/three_geometry_lemmas.pdf}}</ref>

''First proof.'' Let the reflection of {{mvar|AD}} across the angle bisector of {{math|∠''BAC''}} meet {{mvar|BC}} at {{mvar|M'}}. Then:

<math>\frac{|BM'|}{|M'C|} = \frac{|AM'|\frac{\sin\angle{BAM'}}{\sin\angle{ABM'}}}{|AM'|\frac{\sin\angle{CAM'}}{\sin\angle{ACM'}}}
=\frac{\sin\angle{BAM'}}{\sin\angle{ACD}}\frac{\sin\angle{ABD}}{\sin\angle{CAM'}}
=\frac{\sin\angle{CAD}}{\sin\angle{ACD}}\frac{\sin\angle{ABD}}{\sin\angle{BAD}}
=\frac{|CD|}{|AD|}\frac{|AD|}{|BD|}=1</math>

''Second proof.'' Define {{mvar|D'}} as the [[isogonal conjugate]] of {{mvar|D}}. It is easy to see that the reflection of {{mvar|CD}} about the bisector is the line through {{mvar|C}} parallel to {{mvar|AB}}. The same is true for {{mvar|BD}}, and so, {{mvar|ABD'C}} is a parallelogram. {{mvar|AD'}} is clearly the median, because a parallelogram's diagonals bisect each other, and {{mvar|AD}} is its reflection about the bisector.

''Third proof.'' Let {{mvar|ω}} be the circle with center {{mvar|D}} passing through {{mvar|B}} and {{mvar|C}}, and let {{mvar|O}} be the [[circumcenter]] of {{math|△''ABC''}}. Say lines {{mvar|AB, AC}} intersect {{mvar|ω}} at {{mvar|P, Q}}, respectively. Since {{math|1=∠''ABC'' = ∠''AQP''}}, triangles {{math|△''ABC''}} and {{math|△''AQP''}} are similar. Since 
:<math>\angle PBQ = \angle BQC + \angle BAC = \frac{\angle BDC + \angle BOC}{2} = 90^\circ,</math> 
we see that {{mvar|{{overline|PQ}}}} is a diameter of {{mvar|ω}} and hence passes through {{mvar|D}}. Let {{mvar|M}} be the midpoint of {{mvar|{{overline|BC}}}}. Since {{mvar|D}} is the midpoint of {{mvar|{{overline|PQ}}}}, the similarity implies that {{math|1=∠''BAM'' = ∠''QAD''}}, from which the result follows.

''Fourth proof.'' Let {{mvar|S}} be the midpoint of the arc {{mvar|BC}}. {{mvar|1={{abs|BS}} = {{abs|SC}}}}, so {{mvar|AS}} is the angle bisector of {{math|∠''BAC''}}. Let {{mvar|M}} be the midpoint of {{mvar|{{overline|BC}}}}, and It follows that {{mvar|D}} is the [[Inversive geometry#Circle inversion|Inverse]] of {{mvar|M}} with respect to the circumcircle. From that, we know that the circumcircle is an [[Apollonian circles|Apollonian circle]] with [[Focus (geometry)|foci]] {{mvar|M, D}}. So {{mvar|AS}} is the bisector of angle {{math|∠''DAM''}}, and we have achieved our wanted result.

==Tetrahedra==
The concept of a symmedian point extends to (irregular) tetrahedra. Given a tetrahedron {{mvar|ABCD}} two planes {{mvar|P, Q}} through {{mvar|AB}} are isogonal conjugates if they form equal angles with the planes {{mvar|ABC}} and {{mvar|ABD}}. Let {{mvar|M}} be the midpoint of the side {{mvar|{{overline|CD}}}}. The plane containing the side {{mvar|{{overline|AB}}}} that is isogonal to the plane {{mvar|ABM}} is called a symmedian plane of the tetrahedron. The symmedian planes can be shown to intersect at a point, the symmedian point. This is also the point that minimizes the squared distance from the faces of the tetrahedron.<ref name="SBR">{{citation|first1=Jawad|last1=Sadek|first2=Majid|last2=Bani-Yaghoub|first3=Noah|last3=Rhee|title=Isogonal Conjugates in a Tetrahedron|journal=Forum Geometricorum
|volume=16|pages=43–50|year=2016|url=http://forumgeom.fau.edu/FG2016volume16/FG201606.pdf}}.</ref>

==References==
{{reflist}}

==External links==
* [http://www.cut-the-knot.org/Curriculum/Geometry/SymAntiparallel.shtml Symmedian and Antiparallel] at [[cut-the-knot]]
* [http://www.cut-the-knot.org/Curriculum/Geometry/Sym2Antiparallel.shtml Symmedian and 2 Antiparallels] at [[cut-the-knot]]
* [http://www.cut-the-knot.org/Curriculum/Geometry/Symmedian.shtml Symmedian and the Tangents] at [[cut-the-knot]]
* [http://www.uff.br/trianglecenters/X0006.html An interactive Java applet for the symmedian point]
* {{usurped|1=[https://web.archive.org/web/20021223142320/http://www.pballew.net/isogon.html Isogons and Isogonic Symmetry]}}

[[Category:Straight lines defined for a triangle]]