Proof that this quadrilateral is a parallelogram is impossible.

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I found a problem and I think there is a mistake in it because nor just I can't figure out the solution but I also proved by just drawing the figure that the conclusion is wrong. I may also be wrong. Here is the problem:

Ipothesis: Let $ABCD$ a parallelogram and the points $A_1, B_1, C_1$ and $D_1$ in the parallelogram's plan. $M, N, P, Q \in AA_1, BB_1, CC_1, DD_1$ respectively so that $$\frac {MA}{MA_1} = \frac {NB}{NB_1}= \frac {PC}{PC_1} = \frac {QD}{QD_1} = k, k \in \mathbb R \backslash \{1\}$$ Conclusion: If $MNPQ$ is a quadrilateral then $MNPQ$ is parallelogram.

The conclusion sounds strange to me because $MNPQ$ is always a quadrilateral. Am I wrong somewhere?

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A counterexample with random points $A_1,B_1,C_1,D_1$ inside the parallelogram $A,B,C,D$:

enter image description here

points $M,N,P,Q$ forms a quadrilateral, but it is not a parallelogram.

Hence, the hypothesis is false.