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{{ printedBook.courseTrack.name }} {{ printedBook.name }} The midsegment of a trapezoid is parallel to each base and its measure is one half the sum of the lengths of the bases.

If $FE$ is the midsegment of trapezoid $ABCD,$ then $FE∥AB∥DCandFE=21 (AB+CD). $

Let $ABCD$ be a trapezoid and $FE$ be its midsegment.

Draw the line containing the base $AB$ and the line passing through $C$ and $F.$ Let $G$ be the intersection point between these two lines.

The Vertical Angles Theorem gives that $∠AFG≅∠CFD.$ Also, since $GB∥CD,$ then $∠GAF≅∠D$ because of the Alternate Interior Angles Theorem.

With the information written before and the Angle-Side-Angle Congruence Theorem, it can be concluded that $△FAG≅△FDC.$ $△FAG≅△FDC↙↘GA≅CDFG≅FC $
The right-hand side relation above implies that $F$ is the midpoint of $CG.$ In consequence, $FE$ is a midsegment of $△GBC.$ Thus, the Triangle Midsegment Theorem leads to the following conclusion.
$FE∥GBandFE=21 GB $
Because $GB∥AB∥CD,$ then $FE$ is parallel to each base of the trapezoid. Using the Segment Addition Postulate $GB$ can be rewritten as follows.
$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*, it will be obtained the desired result.
$FE=21 GB⇓FE=21 (AB+CD) $