Trapezoid Midsegment Theorem

Let ABCD be a trapezoid and FE be its midsegment. By definition, the midsegment of a trapezoid connects the midpoints of the nonparallel sides.

Next, draw the line connecting the points C and F, then extend the base AB such that it intersects with CF. Let G be the point of intersection between these two lines.

By the Vertical Angles Theorem, it is given that ∠ AFG and ∠ CFD are congruent angles. Also, since GB and CD are parallel, then ∠ GAF and ∠ D are congruent angles by the Alternate Interior Angles Theorem.

Triangles GAF and CDF have two pairs of congruent angles and a pair of included congruent sides. Therefore, by the ASA Congruence Theorem, it is concluded that △ GAF ≅ △ CDF. Since corresponding parts of congruent figures are congruent, then GA≅ CD and FG ≅ FC. △ GAF ≅ △ CDF ↙ ↘ GA≅ CD FG ≅ FC Since FG and FC are congruent, it can be stated that F is the midpoint of CG, and it is given that E is the midpoint of CB. Consequently, FE is a midsegment of △ GBC. Therefore, by the Triangle Midsegment Theorem, FE and GB are parallel and FE is half GB. FE∥ GB and FE = 1/2GB Using the fact that FE and GB are parallel, the first part of the theorem can be obtained.

To obtain the second part, rewrite GB using the Segment Addition Postulate. GB = GA + AB Since GA≅ CD, then GA=CD. GB = CD + AB Finally, substituting the equation above into the equation given by the Triangle Midsegment Theorem, the desired result will be obtained.