How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of length: 2 b c s ( s − a ) b + c I have tried using the sine and cosine rule but have largely failed. A few times I have found a way but they are way too messy to work with.

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How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of length:            2              b        c        s        (        s        −        a        )                    b      +      c      I have tried using the sine and cosine rule but have largely failed. A few times I have found a way but they are way too messy to work with.

reflam2kfnr · Accepted Answer

A method where no trigonometry is used.Consider triangle   A  B  C. Let   A  D, the angle bisector, intersect the circumcircle at   L. Join   L  C. Consider triangle   A  B  D and triangle   A  L  C.Triangle   A  B  D is similar to triangle   A  L  C (by A.A similarity theorem). Therefore,            A      D              A      C        =            A      B              A      L      i.e,  A  D  ⋅  A  L  =  A  C  ⋅  A  B  =  A  D  (  A  D  +  D  L  )  =  A  C  ⋅  A  B  =  A  D  ⋅  A  D  +  A  D  ⋅  D  L  =  A  C  ⋅  A  B   ... (1)By power of point of point result:  A  D  ⋅  D  L  =  B  D  ⋅  D  C  B  D  =  B  C  ⋅            A      B              A      B      +      A      C        D  C  =  B  C  ⋅            A      C              A      B      +      A      C      In (1) ,  A  D  ⋅  A  D  =  A  C  ⋅  A  B  −  B      C    2    ⋅  A  B  ⋅            A      C              (      A      B      +      A      C              )        2              A      D    2    =  A  C  ⋅  A  B            (        1  −            B              C        2                    (      A      B      +      A      C              )        2                        )      If   A  B  =  c  ,  B  C  =  a  ,  A  C  =  b:  A      D    2    =  b  c            (        1  −            a      2              (      b      +      c              )        2                        )

Berghofaei0e · Accepted Answer

Assuming the usage of Trigonometry is allowed,Let   A  D be the bisector of   ∠  B  A  C  △  A  B  C  =      1    2    b  c  sin  ⁡  A  △  A  B  D  =      1    2    ⋅  c  ⋅  A  D  sin  ⁡      A    2   and   △  A  D  C  =      1    2    ⋅  b  ⋅  A  D  sin  ⁡      A    2    △  A  B  C  =  △  A  B  D  +  △  A  D  C  sin  ⁡  A  =  2  sin  ⁡      A    2    cos  ⁡      A    2  As       0    &amp;lt;    A    &amp;lt;    π    ,    0    &amp;lt;          A      2        &amp;lt;          π      2            ⟹        cos    ⁡          A      2        &amp;gt;    0        ⟹        cos    ⁡          A      2        =    +                            1          +          cos          ⁡          A                2             as   cos  ⁡  A  =  2      cos    2    ⁡      A    2    −  1Use       cos    ⁡    A    =                            b          2                +                  c          2                −                  a          2                            2        b        c             and   2  s  =  a  +  b  +  c

How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of l

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