<s>[INST] Prove that only parallelogram satisfies these conditions 
Sum of distances between middle points of two opposite sides of a quadrilateral is equal to its semiperimeter. Prove that the quadrilateral has to be parallelogram.

I have no idea where to start. I tried with using middle line of triangle, messed around with areas, drew parallel lines though vertices, tried some inequalities... 
 [/INST] Hint: Use vector notation. Write out both sides in terms of the position vectors of $A, B, C, D$.
Hint: Apply the triangle inequality to show that equality must occur. 
Hint: Equality only holds when the 3 vertices of a triangle are in a straight line, giving us the parallel condition that we desire.

(Abusing vector notation because I'm lazy to draw the arrows)
Setup: Let the vectors be $A, B, C, D$. The midpoints of 2 opposite sides are $\frac {A+B}{2} $ and $\frac {C+D}{2}$ . So the distance between them is $ | \frac {A+B}{2} - \frac {C+D}{2} | $. Likewise, the other opposite sides are $\frac {A+D}{2} $ and $\frac {B+C}{2}$, so the distance between them is $| \frac {A+D}{2} - \frac {B+C} {2}|$. So, we are given that
$$  | \frac {A+B}{2} - \frac {C+D}{2} | + | \frac {A+D}{2} - \frac {B+C} {2}| = |\frac {A-B}{2} |  + | \frac {B-C} {2} | + | \frac {C-D}{2} | + | \frac {D-A} {2} | $$
Proof: By the triangle inequality, we have
$$ | \frac {A+B}{2} - \frac {C+D}{2} | \leq | \frac {A-D} {2} | + | \frac {B-C}{2} | \\
 | \frac {A+D}{2} - \frac {B+C} {2}| \leq | \frac {A-B}{2} | + | \frac {D-C}{2} | \\$$
Hence, the condition given in the question may only hold when both triangle inequalities hold. For each triangle inequality to hold, we know that all the vectors involved must be parallel to each other. Writing this out, we get
$$ A-D \parallel B-C, A-B \parallel D-C $$
Hence, $A, B, C, D$ define a parallelogram.
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