Let AA', BB', CC', AB", and AC" be cevians of ΔABC with cevians AB" and AC" parallel to lines A'C' and A'B' respectively.

Prove that cevians AA', BB', and CC' are concurrent if and only if A' is the midpoint of line segment B"C".

Note: A cevian is a line segment which joins a vertex of a triangle with a point on the opposite side (or its extension).

AC’’, A’B’ parallel => ACC’’, B’CA’ similar => |CB’|/|B’A| = |CA’|/|A’C’’|      (1)

AB’’, A’C’ parallel => ABB’’, C’BA’ similar => |AC’|/|C’B| = |B’’A’|/|A’B|      (2)

Cevians AA’, BB’, CC’ concurrent

< = >   |CB’|/|B’A| * |AC’|/|C’B| * |BA’|/|A’C|  =  1         (Ceva)

< = >   |CA’|/|A’C’’| *|B’’A’|/|A’B| * |BA’|/|A’C|  =  1       using (1) & (2)

< = >   |B’’A’|/|A’C’’|  =  1

< = >   A’ is the mid-point of B’’C’’

Posted by Harry on 2011-09-21 22:13:02