# 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.

Dec 4, 2015

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

#### Explanation:

Let $A B C D$ be a trapezoid with lower base $A D$ and upper base $B C$.
$M$ is a midpoint of left leg $A B$ and $N$ is a midpoint of right leg $C D$.

Connect vertex $B$ with midpoint $N$ of opposite leg $C D$ and extend it beyond point $N$ to intersect with continuation of lower base $A D$ at point $X$.

Consider two triangles $\Delta B C N$ and $\Delta N D X$. They are congruent by angle-side-angle theorem because
(a) angles ∠$B N C$ and ∠$D N X$ are vertical,
(b) segments $C N$ and $N D$ are congruent (since point $N$ is a midpoint of segment $C D$),
(c) angles ∠$B C N$ and ∠$N D X$ are alternate interior angles with parallel lines $B C$ and $D X$ and transversal $C D$.

Therefore, segments $B C$ and $D X$ are congruent, as well as segments $B N$ and $N X$, which implies that $N$ is a midpoint of segment $B X$.
Now consider triangle $\Delta A B X$. Since $M$ is a midpoint of leg $A B$ by a premise of this theorem and $N$ is a midpoint of segment $B X$, as was just proven, segment $M N$ is a mid-segment of triangle $\Delta A B X$ and, therefore, is parallel to its base $A X$ and equal to its half.

But $A X$ is a sum of lower base $A D$ and segment $D X$, which is congruent to upper base $B C$. Therefore, $M N$ is equal to half of sum of two bases $A D$ and $B C$.

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.