Editing Bisection (section)

====Lengths====

If the side lengths of a triangle are <math>a,b,c</math>, the semiperimeter <math>s=(a+b+c)/2,</math> and A is the angle opposite side <math>a</math>, then the length of the internal bisector of angle A is<ref name=Johnson>Johnson, Roger A., ''Advanced Euclidean Geometry'', Dover Publ., 2007 (orig. 1929).</ref>{{rp|p. 70}}

:<math> \frac{2 \sqrt{bcs(s-a)}}{b+c},</math>

or in trigonometric terms,<ref>Oxman, Victor. "On the existence of triangles with given lengths of one side and two adjacent angle bisectors", ''Forum Geometricorum'' 4, 2004, 215–218. http://forumgeom.fau.edu/FG2004volume4/FG200425.pdf</ref>

:<math>\frac{2bc}{b+c}\cos \frac{A}{2}.  </math>

If the internal bisector of angle A in triangle ABC has length <math>t_a</math> and if this bisector divides the side opposite A into segments of lengths ''m'' and ''n'', then<ref name=Johnson/>{{rp|p.70}}

:<math>t_a^2+mn = bc</math>

where ''b'' and ''c'' are the side lengths opposite vertices B and C; and the side opposite A is divided in the proportion ''b'':''c''.

If the internal bisectors of angles A, B, and C have lengths <math>t_a, t_b,</math> and <math>t_c</math>, then<ref>Simons, Stuart.  ''Mathematical Gazette'' 93, March 2009, 115-116.</ref>

:<math>\frac{(b+c)^2}{bc}t_a^2+ \frac{(c+a)^2}{ca}t_b^2+\frac{(a+b)^2}{ab}t_c^2 = (a+b+c)^2.</math>

No two non-congruent triangles share the same set of three internal angle bisector lengths.<ref>Mironescu, P., and Panaitopol, L.,  "The existence of a triangle with prescribed angle bisector lengths", ''[[American Mathematical Monthly]]'' 101 (1994): 58–60.</ref><ref>[http://forumgeom.fau.edu/FG2008volume8/FG200828.pdf Oxman, Victor, "A purely geometric proof of the uniqueness of a triangle with prescribed angle bisectors", ''Forum Geometricorum'' 8 (2008): 197–200.]</ref>