Let M be a point in the plane of ΔABC such that line AM
bisects ∠BAC. Let A', B', and C' be the perpendicular
projections of point M onto lines BC, CA, and AB respectively.
Let N be the intersection of lines A'M and B'C'.

Prove that line AN bisects side BC.
  

Let E be the intersection of AM with BC. Chasing
angles in right angled triangles: /NMC’ = /ABC
and  /NC’M = /BAM, so triangles NMC’ and EBA
are similar, giving  |MC’||EA] = |NC’||AB|.

By similar reasoning, triangles NMB’ and ECA are
similar, giving  |MB’||EA| = |NB’||AC|.

Since |MB’| = |MC’|,
we now have      |NC’||AB| = |NB’||AC|    (1)

Let PQ denote the position vector of Q relative to P,

AN is parallel to              |NB’| AC’ + |NC’| AB’

so parallel to  (|NB’||AC’|/|AB|) AB + (|NC’||AB’|/|AC|) AC

and, since |AB’| = |AC’|,

AN is parallel to      (|NB’|/|AB|) AB + (|NC’|/|AC|) AC

so parallel to      |NB’||AC| AB + |NC’||AB| AC

Therefore, using (1),    AN is parallel to  AB + AC,

so the line AN passes through the mid-point of BC.

Posted by Harry on 2015-05-18 13:22:12