Prove that the length of the segment connecting the midpoints of the diagonals of a trapezoid equals $\frac 1 2$ the difference in length of its bases (source: Kiselev's Geometry, Vol 1, Ex. 185).
My proof is below. I request verification, critique, suggestions and feedback. Is the proof correct, complete, and rigorous? How could it be improved?
Let $ABCD$ be a trapezoid with midpoint of side $AD$ as $A'$, of side $BC$ as $B'$, of diagonal $BD$ as $P$, and of diagonal $AC$ as $Q$. Since $A'B'$ is the trapezoid's midline, $A'B' = \frac 1 2 DC + \frac 1 2 AB$ and $A'B' \parallel AB \parallel DC$.
Observe that $A'P$ is the midline of $\triangle ABD$ and $QB'$ the midline of $\triangle ABC$, so $A'P = \frac 1 2 AB = QB'$ and $A'P \parallel AB \parallel QB' \parallel A'B'$. Points $A', P, Q, B'$ are therefore colinear; assume WLOG they occur in that order. Consequently:
$$
\begin{align*}
A'P + PQ + QB' = A'B' &= \frac 1 2 DC + \frac 1 2 AB \\
PQ + AB &= \frac 1 2 DC + \frac 1 2 AB \\
PQ &= |\frac 1 2 DC – \frac 1 2 AB|.
\end{align*}$$
Remark: A triangle can be considered a limiting case when $A=B$ and $AB$ vanishes, and a parallelogram when $P=Q$ and $PQ$ vanishes. Both of these still satisfy $PQ = | \frac 1 2 DC – \frac 1 2 AB|$.
Update & Revision
To better understand sirous's succint proof, I rewrote it in my own words. Feedback on correctness and exposition appreciated:
Given trapezoid $ABCD$ with $AB \parallel CD, AB \geq CD$, let $P$ be the midpoint of diagonal $AC$ and $Q$ of diagonal $DB$.
We can then "locate" $P$ by cutting $AB$ at $B'$ such that $AB' \cong DC$. This makes $AB'CD$ a parallelogram with diagonals $AC$ and $DB'$. Since $P$ is the midpoint of $AC$, it is also the midpoint of $DB'$. And since $Q$ is the midpoint of $DB$, then $PQ$ is the midline of $\triangle DBB'$. Thus $$PQ = \frac 1 2 B'B = \frac 1 2 (AB – AB') = \frac 1 2 (AB – DC).$$
Remark: The "crux move" of this proof is to recognize that we need to locate the midpoints of the diagonals, and we can do this by recalling that every trapezoid has an internal parallelogram (really two) which shares one diagonal with the trapezoid. This puts both midpoints on rays of the same vertex, from which the rest follows.
Best Answer
Another approach:
Draw line CG parallel with AB. Quadrilateral ABCG is parallelogra and it's diagonals bisect each other. this means E is mid point ofAC and BG.In triange BGD , EF connects mid point of sides BG and BD and we have:
$EF||GD$ and also $EF=\frac {GD}2$
But:
$AG=BC\Rightarrow GD= AD-(AG=BC)$
Therefore:
$EF=\frac{AD-BC}2$