For the angle bisector I a in a triangle A B C it holds I a 2 = b c ( b + c ) 2 [ ( b + c ) 2 − a 2 ]If I is the incenter, I wonder if there exist similar formula for the part A I 2 .

Question

For the angle bisector       I    a   in a triangle   A  B  C it holds      I    a    2    =            b      c              (      b      +      c              )        2              [  (  b  +  c      )    2    −      a    2    ]If   I is the incenter, I wonder if there exist similar formula for the part   A      I    2  .

Marlee Norman · Accepted Answer

Put weights   a,   b,   c on the three vertices   A,   B,   C. Their center of gravity is   I. The center of gravity of   B and   C is the intersection of   A  I with   B  C.Therefore,   A  I  :      I    a    =                    b        +        c                    a        +        b        +        c

Roland Waters · Accepted Answer

Thank you very much. So the formula is  A      I    2    =            b      c              (      a      +      b      +      c              )        2              [  (  b  +  c      )    2    −      a    2    ]  .

For the angle bisector I a </msub> in a triangle A B C it holds

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