Median of Trapezoid is Parallel to Bases
Theorem
Let $\Box ABCD$ be a trapezoid such that $AB$ and $DC$ are the bases.
Let $E$ be the midpoint of $AD$.
Let $F$ lie on $BC$.
Then:
- $EF$ is parallel to both $AB$ and $DC$
- $F$ is the midpoint of $BC$.
That is, the median of $\Box ABCD$ is parallel to the bases of $\Box ABCD$.
Proof
Sufficient Condition
Let $DH$ be constructed parallel to $BC$ to cut $AB$ at $H$.
From the Parallel Transversal Theorem:
- $DG : GH = DE : EA$
and so $G$ is the midpoint of $AH$.
That is:
- $(1): \quad DG = GH$
Then we have that:
- $DC$ is parallel to $GF$
and:
- $DG$ is parallel to $CF$
so, by definition, $\Box GFCD$ is a parallelogram.
Similarly, we have:
- $GF$ is parallel to $HB$
and:
- $GH$ is parallel to $FB$
so, by definition, $\Box GFBH$ is also a parallelogram.
By Opposite Sides and Angles of Parallelogram are Equal we have that:
- $CF = DG$
and:
- $GH = FB$
But from $(1)$:
- $DG = GH$
and so:
- $CF = FB$
and so $F$ is the midpoint of $CB$.
$\Box$
Necessary Condition
Aiming for a contradiction, suppose $EF$ is not parallel to $DC$.
By Playfair's axiom, there exists a unique straight line through $E$ which is parallel to $DC$.
Let $EF'$ be this line.
From Median of Trapezoid is Parallel to Bases: Sufficient Condition, $F'$ is the midpoint of $BC$.
But by hypothesis $F$ is also the midpoint of $BC$.
That is:
- $F = F'$
So:
- $EF = EF'$
and so $EF$ is parallel to $DC$.
This contradicts our assertion that $EF$ is not parallel to $DC$.
Hence, by Proof by Contradiction, $EF$ is parallel to $DC$.
By definition of base of trapezoid, $AB$ is parallel to $DC$.
By Parallelism is Transitive Relation, $EF$ is parallel to $AB$.
That is, $EF$ is parallel to both $AB$ and $DC$.
$\blacksquare$
Sources
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): median: 3.