Let F be the point on ray AP such that |AF| = |AB|.
We will show that the midpoint of line segment BF
is the fixed point.
Let a, b, and c be the lengths of the sides of ΔABC
opposite vertices A, B, and C respectively and s the
triangles's semiperimeter.
Let line DE intersect line segment BF at point G.
Apply the theorem of Menelaus to ΔBCF and line DE,
|BD| |CE| |FG|
-1 = ------ . ------ . ------
|DC| |EF| |GB|
s - b s - c |FG|
= ------- . ------- . ------
s - c b - s |GB|
or
|FG|
1 = ------
|GB|
Therefore, the point G is fixed and is the midpoint
of line segment BF.
Note: The ratios |XY|/|YZ| in the theorem of
Menelaus are greater than zero if point Y lies
between points X and Z; less than zero otherwise.
See Harry's post for an alternate solution.
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