Theorem
The midpoints of the bases of a trapezoid and the intersection of the diagonals of the same trapezoid are collinear.
Demonstration
$\mathit{ABCD}$ is a trapezoid, and $\mathit{AB}$ and $\mathit{CD}$ are its bases. $O$ is the intersection of the diagonals, and $P$ is the intersection of the lines through the legs of the trapezoid. The line through $O$ and $P$ intersects the bases $\overline{\mathit{AB}}$ and $\overline{\mathit{CD}}$ at $M$ and $N$, respecitvely. $\triangle\mathit{APM} \sim \triangle\mathit{DPN}$, and
\begin{equation*}
\frac{\mathit{AM}}{\mathit{DN}}=\frac{\mathit{PM}}{\mathit{PN}} .
\end{equation*}
$\triangle\mathit{BPM} \sim \triangle\mathit{CPN}$, and
\begin{equation*}
\frac{\mathit{PM}}{\mathit{PN}} = \frac{\mathit{BM}}{\mathit{CN}} .
\end{equation*}
So,
\begin{equation*}
\frac{\mathit{AM}}{\mathit{DN}} = \frac{\mathit{BM}}{\mathit{CN}} .
\end{equation*}
Likewise, since $\triangle\mathit{AOM} \sim \triangle\mathit{CON}$, and $\triangle\mathit{BMO} \sim \triangle\mathit{DNO}$,
\begin{equation*}
\frac{\mathit{AM}}{\mathit{CN}} = \frac{\mathit{BM}}{\mathit{DN}} .
\end{equation*}
I would like to show that the four points $M$, $N$, $O$, and $P$ are collinear. (I will respond to this post with a TikZ
diagram.)
Best Answer
Here is a
TikZ
diagram to accompany the argument provided.