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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 $AD$ intersects with $BC$ at a point. How can I get a contradiction?

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marked as duplicate by YuiTo Cheng, metamorphy, mihaild, воитель, Lee David Chung Lin Jun 6 at 18:32

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  • $\begingroup$ The triangles $OAD,OMN,OBC$ are similar. $\endgroup$ – Yves Daoust Jun 6 at 13:03
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Let $M$ is a midpoint of $AB$, $N$ midpoint of $CD$ and $O$ midpoint of $BD$.

Then $MO=\frac{1}{2}AD$ and $ON=\frac{1}{2}BC$, so $MO+ON=MN$ this means $O$ is located on $MN$, so $MN$ is parallel to $AD$ ($MO$ is a midline of triangle $ABD$) and $MN$ is parallel to $BC$ ($ON$ is a midline of triangle $BCD$). So $BC$ is parallel to $AD$

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I have a different approach: I'll assume we have this quadrilateral and $AD // BC$, where $M$ is the midpoint of $AB$ and $N$ the midpoint of $DC$, then I'll try to prove $MN = \frac{AD+BC}{2}$, or when is this true.

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Consider the triangle of the figure and applying Tales theorem: $$ \frac{OM}{OA} = \frac{MN}{AD} $$ $$ \frac{MN}{BC} = \frac{OM}{OB} $$

Now we solve for $MN$ in the two equations: $$MN = AD \frac{OM}{OA} = AD \left( 1+ \frac{AM}{OA} \right) = AD \left( 1+ x \right),$$ where we have used $OM = OA + AM$ and $x \equiv \frac{AM}{OA}$. And for the second equation: $$MN = BC \frac{OM}{OB} = BC \frac{OA +AM}{OA +2 AM} = BC \frac{1+x}{1+2x},$$ where we have used $AB = 2 AM$.

Now we add the two equations: $$2MN = (1+x)(AD + \frac{BC}{1+2x}),$$ and we recover $$MN = \frac{AD + BC}{2}$$ only when $x = \frac{AM}{OA} = 0$, that is, if $O$ is at $\infty$ and all sides are parallel or we have a parallelogram.

Edit: There must be something wrong because my answer goes in contradiction with @AO1992 answer (that is right), but I can't find the mistake. If someone can point the mistake I will delete the answer.

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