I know how to construct the centroid of a quadrilateral as mentioned here.
But my question is different from that.
We know that if points B,C,G are given in geometry plane and for locating point A such that $G(∆ABC)=G$ we have to do these following things:
(a)Take m(B,C) as shown in figure:
(b) then we reflect m(B,C) in G to get m'(B,C) .
(c) we reflect G in m'(B,C) to get G' and from there we observed that $G(∆G'BC)=G$ so we mentioned $G'=A$.
But if we do such things in case of Quadrilateral then it becomes a challenging task to construct .
So my Question is "If points A,B,C,G are Given then how to locate D such that $G(ABCD)=G$ ?"
My Attempt:
I take $G_b$=$G(∆ABC)$ and placed point $G_d$ on line $GG_b$ then I construct Point D such that $G_d$=$G(∆ADC)$ as shown in figure:
Then Construct point $G_a$ and $G_c$ such that $G_a=G(∆ABD)$ and $G_c=G(∆BCD)$ as shown in figure:
But I observed that Line $G_aG_c$ intersect Line $G_bG_d$ at different points other than G.
So the construction becomes wrong so please help to locate point D such that $G=G(ABCD)$.