An important feature of trapezoids is that the midpoints of its bases, the intersection of its diagonals, and the intersection of the lines through its legs are collinear.
Explanation for three of these points being collinear
$\mathit{ABCD}$ is a trapezoid with bases $\overline{\mathit{AB}}$ and $\overline{\mathit{CD}}$. $M$ is the midpoint of $\overline{\mathit{AB}}$, $N$ is the midpoint of $\overline{\mathit{CD}}$, and $P$ is the intersection of the diagonals. $\angle\mathit{ABP} \cong \angle\mathit{CDP}$, and $\angle\mathit{BAP} \cong \angle\mathit{DCP}$. According to the Angle-Angle Similarity Theorem, $\triangle\mathit{PAB} \sim \triangle\mathit{PCD}$, and \begin{equation*} \frac{\left\vert\overline{\mathit{AB}}\right\vert}{\left\vert\overline{\mathit{CD}}\right\vert} = \frac{\left\vert\overline{\mathit{BP}}\right\vert}{\left\vert\overline{\mathit{DP}}\right\vert}. \end{equation*} According to Apollonius's Theorem, medians $\overline{\mathit{PM}}$ and $\overline{\mathit{PN}}$ are such that \begin{equation*} \frac{\left\vert\overline{\mathit{PM}}\right\vert}{\left\vert\overline{\mathit{PN}}\right\vert} = \frac{\left\vert\overline{\mathit{AB}}\right\vert}{\left\vert\overline{\mathit{CD}}\right\vert} = \frac{\left\vert\overline{\mathit{BM}}\right\vert}{\left\vert\overline{\mathit{DN}}\right\vert}. \end{equation*} So, \begin{equation*} \frac{\left\vert\overline{\mathit{BM}}\right\vert}{\left\vert\overline{\mathit{DN}}\right\vert} = \frac{\left\vert\overline{\mathit{BP}}\right\vert}{\left\vert\overline{\mathit{DP}}\right\vert} = \frac{\left\vert\overline{\mathit{PM}}\right\vert}{\left\vert\overline{\mathit{PN}}\right\vert} , \end{equation*} and, according to the Side-Side-Side Similarity Theorem, $\triangle\mathit{BMP} \sim \triangle\mathit{DNP}$. In partiuclar, $\angle\mathit{BPM} \cong \angle\mathit{DPN}$. Consequently, \begin{align*} \angle\mathit{MPN} &= \angle\mathit{BPM} + \angle\mathit{BPN} \\ &= \angle\mathit{BPM} + (180^\circ - \angle\mathit{DPN}) \\ &= \angle\mathit{BPM} + (180^\circ - \angle\mathit{BPM}) \\ &= 180^\circ. \end{align*} Consequently, $M$, $N$, and $P$ are collinear.
Explanation for another three points being collinear
If $Q$ is the intersection of the lines through the legs of the trapezoid, I would like an explanation for the collinearity of $M$, $N$, and $Q$ - an elementary explanation like the one showing the collinearity of $M$, $N$, and $P$.
Additionally, a careful explanation using a homothety would be much appreciated. I am starting to familiarize myself with the concept, and this would be a nice illustration.
Consider line $QM$ and let $N'$ the point where it meets $CD$. Triangles $QDN'$ and $QAM$ are similar (or homothetic, if you prefer), hence $DN'/AM=QN'/QM$. Triangles $QCN'$ and $QBM$ are also homothetic, hence $CN'/BM=QN'/QM$. It follows that $DN'/AM=CN'/BM$, and from $AM=BM$ we have then: $DN'=CN'$, meaning that $N'=N$ and concluding the proof.