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|

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|

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