Let A', B', and C' be the midpoints of the sides opposite
vertices A, B, and C respectively. If P equals A, B, or C',
then the desired line is AA', BB', or CC' respectively.
If P lies between B and C', then
the desired line must intersect side AC. Construct a
line through B parallel to line PB' that intersects side
AC at point Q. Line PQ is the desired line. Proof:
From similar triangles APB' and ABQ,
|AP| |AB'|
------ = ------- or
|AB| |AQ|
|AP||AQ| = |AB||AB'| or
|AP||AQ| = |AB||AC|/2 or
(1/2)|AP||AQ|sin(A) = (1/2)|AB||AC|sin(A)/2 or
Area(APQ) = Area(ABC)/2
else if P lies between A and C', then
same as above swapping letters "A" and "B".
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