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Let $ABCD$ be a convex quadrilateral such that the length of the segment connecting midpoints of the two opposite sides $AB$ and $CD$ equals $\frac{AD+BC}{2}$ . Prove that AD is parallel to BC.
I assume that AD is not parallel to BC but I can't find any contradiction.

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Lines with the same color code are parallel.

We are assuming AD is not parallel to BC. enter image description here

Let our target line be MN. Through A draw AP parallel to MN cutting CD at P. Q is similarly constructed.

After joining AQ, we get MN = $\dfrac { BQ + AP }{2}$.

Form the parallelograms PAQY and PACZ. Then CZYQ is also a parallelogram.

Note that N is the midpoint of both CD and QP. This means DP = QC = YZ.

By SAS, $\triangle CZY \cong \triangle APD$. This means AD = CY.

On one hand, we have MN = $\dfrac { BQ + AP }{2} = \dfrac { BQ + QY }{2} = \dfrac {BY}{2}$.

On the other hand, according to the given, MN = $\dfrac {BC + AD}{2} = \dfrac {BC + CY}{2}$.

But BC + CY > BY according to the triangular inequality. Hence, we have a contradiction.

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