Let D be the intersection of line AM and line B'C'.
Let E be the intersection of line AM and side BC.
Let F be the midpoint of side BC.
PRELIMINARY:
(a) ∠ABE = ∠C'MN because BA'MC' is a cyclic
quadrilateral and the interior
angle at a vertex is equal to
the exterior angle at the
opposite vertex.
(b) ∠BAE = ∠MC'N because the right triangles
ADC' and C'DM are similar.
(c) (a) & (b) ==> ΔC'MN ~ ΔABE by ASA.
==> |NC'||AB| = |MC'||AE|
(d) ∠ACE = ∠B'MN because CB'MA' is a cyclic
quadrilateral and the interior
angle at a vertex is equal to
the exterior angle at the
opposite vertex.
(e) ∠CAE = ∠MB'N because the right triangles
ADB' and B'DM are similar.
(f) (d) & (e) ==> ΔB'MN ~ ΔACE by ASA.
==> |NB'||AC| = |MB'||AE|
|NC'| |NB'|
(g) (c) & (f) ==> ------- = -------
|AC| |AB|
because |MB'| = |MC'|.
PROOF:
Let PQ denote the vector from point P to point Q.
Let s = +1 if C' lies on ray AB
= -1 if not.
AN = AC' + C'N
|C'N|
= AC' + -------- C'B'
|C'B'|
|NC'|
= AC' + --------*( AB' - AC')
|B'C'|
|NC'| |NC'|
= ( 1 - -------- ) AC' + -------- AB'
|B'C'| |B'C'|
|NB'| |NC'|
= -------- AC' + -------- AB'
|B'C'| |B'C'|
|NB'|*s*|AC'| |NC'|*s*|AB'|
= --------------- AB + --------------- AC
|B'C'||AB| |B'C'||AC|
Because |AC'| = |AB'| and the
PRELIMINARY
|NB'|*s*|AB'|
= ---------------*( AB + AC )
|B'C'||AB|
2*|NB'|*s*|AB'|
= -----------------*( AB + AC )/2
|B'C'||AB|
2*|NB'|*s*|AB'|
= -----------------*AF
|B'C'||AB|
Therefore, since line AN coincides with vector AN,
line AN bisects side BC.
QED
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