$\displaystyle \setlength{\unitlength}{1.3mm}
\begin{picture}(50,45)
\put(24.6,38){$\bullet$}
\put(19.6,5.6){$\bullet$}
\put(33.7,3.9){$\bullet$}
\put(21.6,39){$A$}
\put(0,6){$B$}
\put(36,3.2){$C$}
\put(24,16){$O$}
\put(20,3){$D$}
\put(36,22.5){$X$}
\put(43,11){$Y$}
\put(25.6,38.6){\line(3,-5){20}}
\put(45,11){\line(-5,-3){20}}
\thicklines
\put(39,0){\line(-1,1){34}}
\put(26,41){\line(-1,-6){7}}
\put(42,29){\line(-5,-3){42}}
\end{picture}$
Given lines AO, BO, CO, you want to locate the positions of A, B, C on those lines so that they form the vertices of a triangle having those lines as medians. The size of the triangle is not determined by that condition, so you can fix A arbitrarily on the line AO. Construct the point D on the line AO, on the opposite side of O from A, so that $\displaystyle DO = \tfrac12AO$. Then D will lie on the side BC of the triangle.
The centroid O of the triangle is the centre of mass of three equal masses at the vertices. So the vertices A and C must be equidistant from the median BO. Construct a line AXY through A, perpendicular to BO, meeting BO at X, and such that AX = XY. Then construct a line through Y perpendicular to AXY (and therefore parallel to BOX). This line meets CO at the vertex C of the triangle, and thus determines the position of C.
Finally, you can construct B as the point where the line CD meets the median BOX.