What is the simplest solution to this elementary geometry problem? The Problem
Given a triangle $ABC$, points $M$ and $N$ are chosen in such way that the midpoint of segment $BM$ is coincident with the midpoint of side $AC$, and the midpoint of segment $CN$ is coincident with the midpoint of side $AB$. Prove that points $M$, $N$, $A$ are collinear.
The drawing is provided.

The Solution
Below I provide my own solution of the stated problem.
Let $P$ be the midpoint of $AB$, and $Q$ the midpoint of $AC$. We are going to prove that $A\in MN$.

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*$\overline{CP}=\overline{PN}$, $\overline{AP}=\overline{PB}$ (by statement of the problem), $\angle CPB = \angle APN$ (these angles are vertical by definition) $\Rightarrow _\Delta BPC\cong _\Delta APN$ (SAS congruence).


*$_\Delta BPC\cong _\Delta APN\Rightarrow \angle PCB=\angle PNA\Rightarrow BC\parallel AN$ (from the congruence of alternate angles).


*$\overline{BQ}=\overline{QM}$, $\overline{AQ}=\overline{QC}$ (by statement of the problem), $\angle BQC = \angle AQM$ (these angles are vertical by definition) $\Rightarrow _\Delta BQC\cong _\Delta MQA$ (SAS congruence).


*$_\Delta BQC\cong _\Delta MQA\Rightarrow \angle CBQ=\angle AMQ\Rightarrow BC\parallel AM$ (from the congruence of alternate angles).


*$BC\parallel AN$ and $BC\parallel AM$, so $BC$ is parallel to two line segments with a common endpoint. Clearly, as $ABC$ is a triangle, point $A$ is not on line $BC$. But by Playfair's axiom, at most one line parallel to the given line can be drawn through the point not on this line. Thus, two segments with a common endpoint can be parallel to the same line if and only if they are on the same line. So we conclude that $\overline{AM}$ and $\overline{AN}$ are collinear, and thus, $A$, $M$, $N$ are collinear, QED.
The Question
I'm looking for the simplest solution of this problem you can find. By simplest I mean the most elementary, using only basic facts of geometry, preferrably more elementary than the one I came up with above.
 A: $ACBN$ and $ABCM$ are parallelograms, because their diagonals are bisected by the point of intersection. So, $AN \parallel BC$ and $AM \parallel BC$.
A: Let $X$ be the intersection of $BM$ and $AC$, and similarly for $CN$ and $AB$ with $Y$.
Rotate $\triangle BCX$ by $180^{\circ}$ about $X$ (and the same with $\triangle BCY$ about $Y$) to get the result.
A: As JMP does, we'll call $ \ X \ $ the midpoint of $ \ \overline{AC} \ $ and $ \ Y \ $ the midpoint of $ \ \overline{AB} \ \ . \ $  Then $ \ \overline{BX} \ $ and $ \ \overline{CY} \ $ are medians of $ \ \Delta ABC \ \ . \ $  These intersect at the centroid $ \ P \ $ of the triangle and thus  $ \ \overline{BP} \ $ and $ \ \overline{CP} \ $ have $ \ \frac23 \ $ of the length of  $ \ \overline{BX} \ $ and $ \ \overline{CY} \ \ , \ $  respectively.  Hence,   $ \ \overline{MP} \ $ and $ \ \overline{NP} \ \   $  have $ \ \frac43 \ $ the length of   $ \ \overline{BX} \ $ and $ \ \overline{CY} \ \ , \ $  respectively, making triangles $ \ \Delta MPN \ $ and $ \ \Delta BPC \ $ similar (the two triangles having vertical angles at $ \ P \ ) \ . $ So $ \ \overline{MN} \ $ and $ \ \overline{BC} \ $ are parallel, as we have pairs of alternate interior angles among the two triangles.
If we call $ \ Q \ $ the midpoint of $ \ \overline{BC} \ \ , \ $ then $ \ \overline{AQ} \ $ is also a median of  $ \ \Delta ABC \ \ . $  The similarity of $ \ \Delta MPN \ $ and $ \ \Delta BPC \ $ places $ \ A \ $ on the line $ \ \overline{MN} \ \ . $
[The proposition is very simple to prove analytically:  placing $ \ A \ $ at the origin (and $ \ B \ $ and $ \ C \ $ at arbitrary coordinates) and finding the coordinates of the midpoints $ \ X \ $ and $ \ Y \ \ , \ $ and thence those of $ \ M \ $ and $ \ N \ \ , \ $ it is straightforward to show that the slope of side $ \ \overline{BC} \ $ is the same as the line containing $ \ \overline{MN} \ \ . \ $ It is then found that the equation of said line passes through the origin.]
