For a convex quadrilateral $ABCD$, let $K$,$L$,$M$,$N$,$P$,$Q$ be the midpoints of $AB$,$BC$,$CD$,$DA$,$AC$,$BD$ $\square QNPE$, $\square KLMN$ are parallelogram and three line of the problem are concurrent at their common midpoint.

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Obel1x wrote: Prove that the lines joining the middle-points of non-adjacent sides of an convex quadrilateral and the line joining the middle-points of diagonals, are concurrent. Prove that the intersection point is the middle point of the three given segments. Let $ABCD$ be the convex quadrilateral, and other points defined as in the picture. Since $P$ is the midpoint of $AB$, and $R$ is the midpoint of $CD$ then the midpoint of the segment $PR$ is $T=\left(\frac{x_a +x_b +x_c+x_d}{4}, \frac{y_a +y_b +y_c +y_d}{4}\right)$. Similarly, the midpoint of $QS$ has coordinates $\left(\frac{x_a +x_b +x_c+x_d}{4}, \frac{y_a +y_b +y_c +y_d}{4} \right)$; the coordinates of the midpoint of the segment $XY$ are $\left (\frac{x_a +x_b +x_c +x_d}{4}, \frac{y_a +y_b +y_c +y_d}{4}\right)$. Therefore segments $PR, QS$ and $XY$ have their midpoint in common. This solves the problem. [geogebra]049e3526058a51bc409235ddc1ed9a5b5211f9f1[/geogebra]