Let points L, M, and N lie on the sidelines BC, CA, and AB
of ΔABC respectively. If the lines AL, BM, and CN are
concurrent at point P. Prove

      AP     AN     AM
     ---- = ---- + ----
      PL     NB     MC 

, where the six line segments are directed and
the three denominators are not zero.
  

Triangles APC and BPC have the same base, PC, so the ratio of their
areas is given by the ratio of their perpendicular heights and therefore
also by |AN|/|NB|. Taking account of directions, and using square
brackets to represent area (with the order of vertices considered),
we can write this as:

AN/NB = [APC]/[ PBC]    and, similarly,    AM/MC = [ABP]/[PBC]

Thus:           AN/NB +  AM/MC    =  ([APC] + [ABP])/[PBC]
                                                =  ([ABC] – [PBC])/[PBC]
                                                =  [ABC]/[PBC]  -  1
                                                =  AL/PL  -  1
                                                =  AP/PL

Posted by Harry on 2014-06-02 14:46:07