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.


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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