Let ABCD be a trapezoid with lower base AD and upper base BC.
M is a midpoint of left leg AB and N is a midpoint of right leg CD.
Connect vertex B with midpoint N of opposite leg CD and extend it beyond point N to intersect with continuation of lower base AD at point X.
Consider two triangles Delta BCN and Delta NDX. They are congruent by angle-side-angle theorem because
(a) angles ∠BNC and ∠DNX are vertical,
(b) segments CN and ND are congruent (since point N is a midpoint of segment CD),
(c) angles ∠BCN and ∠NDX are alternate interior angles with parallel lines BC and DX and transversal CD.
Therefore, segments BC and DX are congruent, as well as segments BN and NX, which implies that N is a midpoint of segment BX.
Now consider triangle Delta ABX. Since M is a midpoint of leg AB by a premise of this theorem and N is a midpoint of segment BX, as was just proven, segment MN is a mid-segment of triangle Delta ABX and, therefore, is parallel to its base AX and equal to its half.
But AX is a sum of lower base AD and segment DX, which is congruent to upper base BC. Therefore, MN is equal to half of sum of two bases AD and BC.
End of proof.
The lecture dedicated to this and other properties of quadrilaterals as well as many other topics are addressed by a course of advanced math for high school students at Unizor.