# Trapezoid and inscribed square

I need help with this hard geometry problem.

Consider a trapezoid ABCD (AB||CD and AD=BC). A square $A_1A_2A_3A_4$ is inscribed in the trapezoid such that $A_1 \in AB$, $A_2\in BC$, $A_3\in CD$ and $A_4 \in AD$. If $M$ and $N$ are the midpoints of AD and BC and $A_1A_3 \cap A_2A_4 = O$, prove that $O \in MN$.

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One vertex ($A_3$) of the inscribed square touches the top edge, another ($A_1$) the bottom. $O$ lies on the middle of the line segment $A_1A_3$; hence it lies at half the height of the trapezoid.