The proof is explained below:

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 ΔBCN and ΔNDX . 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 ΔABX . 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 ΔABX 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

You can also go for youtube, for better understanding through video lectures. I hope this helps you!