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The straight dope on midlines

Okay let's do some math.

The midline of a trapezoid is the line connecting the midpoints of the non-parallel sides

Yup! There's the midline.

Great! You guess pretty well! In fact, it's always true that the length of the midline of a trapezoid is equal to the average of the lengths of the opposite sides.

Here's a sketch of a proof if you're interested. I intentionally left out a few details for you to fill in :).

Take $X$ on $BC$ such that $DX||AB$ and $Y$ to be the midpoint of $DX$.

The rest of the proof goes like this: First we show $E,Y,F$ are collinear (even though it's obvious, our definition of $Y$ doesn't give it to us for free), and then we crank out some computations.

By SAS similarity, $\triangle DEY\sim \triangle DCX$ so $EY||BC$ and $|EY|=\frac{|CX|}{2}$. Since $ABXD$ is a parallelogram, $ADYF$ and $BXYF$ are also parallelograms (why?), so $BX||YF||AD$ and $|BX|=|YF|=|AD|$ (why?).

Now since $EY$ and $FY$ are both parallel to $BC$, we get $E,Y,F$ are collinear.

Thus, $$\begin{align} |EF| &= |EY|+|YF| \\ &= \frac{|CX|}{2} + |AD| \\ &= \frac{|BC|-|AD|}{2} + |AD| \\ &= \frac{|BC|+|AD|}{2}. \end{align}$$

Great job! Now you know what the midline is, and a cool property of it! Hope you like trapezoids. :)

Questions 7-12 on this worksheet are about midlines of trapezoids, which I'm sure you love by now, thanks to Ray! Have fun solving them!