NOTATION:
Let a, b, and m be the side lengths opposite vertices A, B, and M respectively.
Let X' be the endpoint of the cevian extended from vertex X.
CONSTRUCTION:
Line segment BM determines vertices B and M. Vertex A lies on the
circle with center B and radius |AB|. If point A' on line segment BM
can be constructed, then the line through A' and perpendicular to BM
will intersect the circle at vertex A.
PROOF:
Let x = |BA'|. Then the cevians are concurrent by Ceva's theorem if
|AM'| |BA'| |MB'|
------- • ------- • ------- = 1
|M'B| |A'M| |B'A|
or
x a
1 • ----- • --- = 1
a-x m
or
a*m
x = -----
a+m
Length x is easily constructed from lengths a and m.
QED
NOTE:
See Harry's post for an alternate solution.
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