Question

How do you prove this theorem on trapezoids and its median? The median (or mid-segment) of a trapezoid is parallel to each base and its length is one half the sum of the lengths of the bases.

Answer 1

The proof of this theorem about mid-segment of a trapezoid is below.

Explanation:

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.