Consider a point $X$ inside $\triangle ABC$, and construct $AD, BE, CF$ passing through $X$ and ending on the opposite sides. Let these be vectors. Prove that if $\vec{AD}+\vec{BE}+\vec{CF}$ is equal to $0$ then $X$ is the centroid.
Because this involves vectors, I am thinking a potential approach would use barycentric coordinates, and to take advantage of the properties of $D,E,F$ determined by the center. How would I prove this?
Edit : So I have barycentrics for the vectors $AD, BE, CF$ as coordinates where $a+b=1, c+d=1, e+f=1$ and $(1,-a,-b), (-c,1,-d), (-e,-f,1)$. How do I relate themselves to a common center?
Let the coordinates be $A=(1,0,0), B=(0,1,0), C=(0,0,1)$, $D=(0,t,1-t), E=(1-u,0,u), F=(v,1-v,0)$.
Writing $\vec{AD}=\vec{D}-\vec{A}$ etc, the condition $\vec{AD}+\vec{BE}+\vec{CF}=0$ gives $$\vec{D}+\vec{E}+\vec{F}=\vec{A}+\vec{B}+\vec{C}$$ In terms of the barycentric coordinates, $$(1-u+v, t+1-v, 1-t+u)=(1,1,1)$$ Equating the corresponding coordinates yields $t=u=v$. Finally for concurrence to hold, by Ceva's theorem, $tuv = (1-t)(1-u)(1-v)$. Setting $u=v=t$ we have $t^3=(1-t)^3\Rightarrow t=1-t \Rightarrow t=1/2$.
Thus $D,E,F$ are midpoints of the sides and the concurrence point $X$ is indeed the centroid of $\triangle ABC$.