Editing Symmedian (section)

== 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.