Let $ABCD$ be a trapezoid and $\overline{FE}$ 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 $\overline{AB}$ such that it intersects with $\overleftrightarrow{CF}.$ Let $G$ be the point of intersection between these two lines.

By the Vertical Angles Theorem, it is given that $\angle AFG$ and $\angle CFD$ are congruent angles. Also, since $\overline{GB}$ and $\overline{CD}$ are parallel, then $\angle GAF$ and $\angle D$ 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 $△GAF\cong △CDF.$ Since corresponding parts of congruent figures are congruent, then $\overline{GA}\cong \overline{CD}$ and $\overline{FG}\cong \overline{FC}.$ Since $\overline{FG}$ and $\overline{FC}$ are congruent, it can be stated that $F$ is the midpoint of $\overline{CG},$ and it is given that $E$ is the midpoint of $\overline{CB}.$ Consequently, $\overline{FE}$ is a midsegment of $△GBC.$ Therefore, by the Triangle Midsegment Theorem, $\overline{FE}$ and $\overline{GB}$ are parallel and $FE$ is half $GB.$ Using the fact that $\overline{FE}$ and $\overline{GB}$ are parallel, the first part of the theorem can be obtained. To obtain the second part, rewrite $GB$ using the Segment Addition Postulate. Since $\overline{GA}\cong \overline{CD},$ then $GA=CD.$ Finally, substituting the equation above into the equation given by the Triangle Midsegment Theorem, the desired result will be obtained.