Equilateral and Stewart

Consider an equilateral ABC and the point E on side BC dividing BC in ratio BE/EC = 1/3. Find point F on side AB which is equidistant from A and E. Show that  |FB|+|BE| = |FC| ([Court, p. 32]).
The exercise is interesting since it gives the occasion to apply Stewart in a simplified form and deduce several relations. To write some of them introduce s = |AB|, for the side of the equilateral, x = |AF|, y = |AN|, z = |BN|.
1) x = |AF| = (7/15)s,
2) y = |AN| =  (7/8)s,
3) z = |BN| = (5/8)s.
These imply that |CF| = (13/15)s, and the proof of the statement |FB|+|BE| = |FC|.

Project A parallel to FE to N on side BC. From the similar triangles BEF and BNA:
1) |BN|/|AB| = |BE|/|BF| => (z/s) = (s/3)/(s-x).
2) |BN|/|BE| = |AN|/|EF| => (z/(s/3)) = y/x.
From Stewart's theorem we have for the length of AN:
3) |AN|2 = s2(1-(z/s)((s-z)/s).
The three equations can be easily solved for x, y, z and deliver the stated values. The proof of the statement follows by applying Stewart again to the sevian CF:
|CF|2 = s2(1-(x(s-x))/s2) = s2 (1-(56/152)) => |CF| = (13/15)s.

Bibliography

[Court] Altshiller-Court, Nathan College Geometry: A Second Course in Plane Geometry for Colleges and Normal Schools, 2nd Ed.. New York, Barnes and Noble, 1952