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In trapezoid ABCD, point E is the intersection point of the diagonals.
A segment parallel to the bases was drawn through point E and it intersects the sides of the trapezoid in points F and G. Prove that EF = FG.

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What is the easiest and the most intuitive way to solve that? I know there are lots of them, but many are too abstract for someone unversed in geometry.

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    Duplicate of (http://math.stackexchange.com/q/141701) – 2017-02-13
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    An interesting connection with the harmonic mean:(http://jwilson.coe.uga.edu/emt725/Isos.Trpzd/Diag/diag.html) – 2017-02-13

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I think you want FE = EG instead.

$\triangle DFE \sim \triangle DAB$. Get the other similar pair. They both have the same base AB. Intercept theorem is needed to bridge the result.