# A geometry question of quadrilateral. [duplicate]

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?

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

• The triangles $OAD,OMN,OBC$ are similar. – Yves Daoust Jun 6 at 13:03

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 $$AD$$

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.

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.