Lemma: In triangle $ABC$, $D, E, F$ are points on sides $BC, CA, AB$ respectively. Triangles $ABC$ and $DEF$ share the same centroid $G$. $P, Q, R, D'$ are the midpoints of $BC, CA, AB, EF$ respectively. Then $Q, D', R$ are collinear. Proof: Construct a point $M$ such that $AMPD$ is a parallelogram. Now, $G$ is the intersection point of $AP$ and $DD'$ because $G$ is common for both the triangles $ABC$ and $DEF$. In triangles $ADG$ and $PD'G$, we have $\angle AGD=\angle PGD'$ and $\frac{AG}{GP}=\frac{DG}{GD'}$ by properties of centroid $G$ for both the triangles $ABC$ and $DEF$. So, we have $\Delta ADG\sim \Delta PD'G$. Note that this implies $\angle GAD=\angle GPD'$ and hence $AD$ is parallel to $D'P$. We also have $AD=2D'P$. But since $AMPD$ is a parallelogram, $AD=MP=2D'P$ and so, $D'$ is the midpoint of $MP$. Now, if we draw a parallel to $BC$ passing through $D'$, let it meet $AD$ at $D''$ and $AC$ at $Q'$ and $AB$ at $R'$. Since $AM$ is parallel to $D'P$ which is in turn parallel to $D'D''$, we have $D''$ to be the midpoint of $AD$ in parallelogram $AMPD$. Now, in triangle $ADC$, since $D''Q'$ is parallel to $DC$ and $D''$ is the midpoint of $AD$, by midpoint theorem, we have $Q'$ to the midpoint of $AC$. Or in other words $Q'$ coincides with $Q$. In triangle $ADB$, since $D''$ is the midpoint of $AD$ and $D''R'$ is paralell to $AB$, we have $R'$ to be the midpoint of $AB$ by midpoint theorem and hence it coincides with $R$. Since $R', D', Q'$ are collinear, this is same as $Q, D', R$ to be collinear. And hence, we have proved our lemma. Now, coming back to the problem. Let $P, Q, R$ be the midpoints of $BC, CA, AB$ respectively and $D', E', F'$ be the midpoint of $EF, FD, DE$ respectively. So, from our lemma, the set of points $(P, F', Q), (Q, D', R), (R, E', P)$ are collinear. Note that $P, Q$ are midpoints of $BC, CA$ respectively and by converse of midpoint theorem, $PQ$ is parallel to $AB$ and hence $PF'$ is parallel to $AB$. Now,we draw a parallel to $AB$ passing through $E$ meeting $BC$ at $D''$. We have $ED''$ to be parallel to $BA$ which is in turn parallel to $PF'$. In triangle $DED''$, since $F'P$ is parallel to $ED''$ and $F'$ is the midpoint of $DE$ and by converse of midpoint theorem, $P$ is the midpoint of $DD''$. Since $P$ is also the midpoint of $BC,$ we have $BD=CD''$ and $DC=D''B$. So, $\frac{BD}{DC}=\frac{CD''}{D''B}$. We also have $ED''$ to be parallel to $AB$ by construction, so by basic proportionality theorem, $\frac{CD''}{D''B}=\frac{CE}{EA}$ and hence, $\frac{BD}{DC}=\frac{CE}{EA}$. Due to cyclic symmetry, we have \[ \frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB} \] For the only if condition, I think retracing the steps would be possible.

Let $M,N$ be the midpointsof $FE,BC$, $AN\cap BC=D'$. The two triangles have the same center $\iff AM,DN$ trisect each other i.e,if $AM\cap DN=G$, then $AG=2GM,DG=2GN$.$\iff N,M$ are the midpoints of $AD,DD'$ resp. $\iff AED'F$ is a parallelogram $\iff \frac {BD}{DC}=\frac {BD'}{CD'}=\frac {CE}{AE}=\frac {AF}{BF}$ and we are done.

Let $M$ be the middle point of $BC$, $P$ the middle point of $DE$, and $G$ be the intersection of $AM$ and $FP$. Consider $D'$ a point on $BC$ such that ${{CD'}/{D'B}}={{BD}/{DC}}$. It follows that \[\frac{BD}{BC} = \frac{CE}{CA}=\frac{AF}{AB}=\frac{CD'}{CB} \hspace{.5 cm}(1)\][asy][asy] size(200); defaultpen(fontsize(10)); pair B=origin, C=(1,0), A=(0.25,0.5), point=(1.25/3.0,0.5/3.0); pair D=waypoint(B--C,0.3); pair E=waypoint(C--A,0.3); pair F=waypoint(A--B,0.3); pair D1=waypoint(B--C,0.7); pair E1=waypoint(C--A,0.7); pair F1=waypoint(A--B,0.7); pair M=waypoint(B--C,0.5); pair N=waypoint(C--A,0.5); pair O=waypoint(A--B,0.5); pair P=waypoint(D--E,0.5); pair G= intersectionpoint(A--M,F--P); draw(B--C--A--B, linewidth(0.7)); draw(A--M, linewidth(0.5)); draw(D--E, linewidth(0.5)); draw(F--P, linewidth(0.5)); draw(M--P,dotted+linewidth(0.5)); draw(E--D1,dotted+linewidth(0.5)); //dot(A^^B^^C^^D^^E^^F^^D1^^E1^^F1^^M^^N^^O^^P^^G); dot(A^^B^^C^^D^^E^^F^^D1^^M^^P^^G); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D-(0,0.01), dir(point--D)); label("$E$", E, dir((point-(.5,.5))--E)); label("$F$", F, dir(point--F)); label("$D'$", D1-(0,0.015), dir(point--D1)); //label("$E'$", E1, dir((point)--E1)); //label("$F'$", F1, dir(point--F1)); label("$M$", M, dir(point--M)); //label("$N$", NE, dir(point--N)); //label("$O$", O, dir(point--O)); label("$P$", P, NE); label("$G$", G, E); [/asy][/asy] Now, ${{CD'}\over{CB}}={{CE}\over{CA}} \Rightarrow {ED'}||{AB}\hspace{.25 cm}(*), {{ED'}\over{AB}}={{CE}\over{CA}}\hspace{.25 cm}(**)$. $((1), (**)) \Rightarrow ED'=AF \hspace{0.5 cm} (2)$. $(CD' = BD, BM = CM) \Rightarrow MD=MD' \hspace{0.5 cm} (3)$ $(P$ is the middle point of $DE$, $(3)) \Rightarrow (PM||ED', ED' = 2PM) \hspace{0.5 cm} (4)$ $((4), (*), (2)) \Rightarrow (PM||AF, AF = 2PM) \hspace{0.5cm} (5)$ $(5) \Rightarrow \Delta{AFG} ~ \Delta{GPM} \Rightarrow \frac{AG}{GM} = \frac{FG}{GP}=\frac{AF}{PM}=2$ The fact that $G$ divides the medians $AM$ and $FP$ with a ratio of $2:1$ means that $G$ is the centroid of $\Delta{ABC}$ and $\Delta{DFE}$. The proof of the reverse is straight forward as we can reverse all the above steps to show that $\frac{CE}{CA}=\frac{AF}{AB}.$ By considering the medians of $\Delta{ABC}$ and $\Delta{DFE}$ coming out of $B$ and $D$ we can show $\frac{AF}{AB}=\frac{BD}{DC}.$ From these two equalities we can show $\frac{BD}{DC} = \frac{CE}{EA}=\frac{AF}{FB}.$

With barycentric coordinates, $A=\left(1,\ 0,\ 0\right),\ B=\left(0,\ 1,\ 0\right),\ C=\left(0,\ 0,\ 1\right),\ G=\left(\frac{1}{3},\ \frac{1}{3},\ \frac{1}{3}\right)$ Let $D=\left(x,\ 1-x,\ 0\right),\ E=\left(0,\ y,\ 1-y\right),\ F=\left(1-z,\ 0,\ z\right)$ so the centroid of $DEF$ is $G'=\left(\frac{1+x-z}{3},\ \frac{1+y-x}{3},\ \frac{1+z-y}{3}\right)$ If $G=G'$ then $\frac{1+x-z}{3}=\frac{1}{3},\ \frac{1+y-x}{3}=\frac{1}{3},\ \frac{1+z-y}{3}=\frac{1}{3}$ which implies that $x=y=z$ which is analogous to what we need to prove.

Amir Hossein wrote: The points $D,E$ and $F$ are chosen on the sides $BC,AC$ and $AB$ of triangle $ABC$, respectively. Prove that triangles $ABC$ and $DEF$ have the same centroid if and only if \[\frac{BD}{DC} = \frac{CE}{EA}=\frac{AF}{FB}\] This is pretty much obvious using complex numbers....

H.HAFEZI2000 wrote: This is pretty much obvious using complex numbers.... Your comment doesn't help anyone. If it's obvious, post your solution.

Amir Hossein wrote: H.HAFEZI2000 wrote: This is pretty much obvious using complex numbers.... Your comment doesn't help anyone. If it's obvious, post your solution. Either I solved this incorrectly, or the proof really is "obvious" as @above pointed out (even though I hate that word). In the language of complex numbers, $ABC$ and $DEF$ have the same centroid iff $\frac{1}{3}(a+b+c)=\frac{1}{3}(d+e+f)$. The ratio $\frac{BD}{DC}$, which we assume is real when described in complex coordinates given that $B,D,C$ are collinear, is given by: $$\frac{b-d}{d-c}=\frac{(e+f)-(a+c)}{(a+c)-(e+f)}=-1$$. Of course as all points are defined analogously, and $-1$ is a real constant, $\frac {CE}{EA}$ and $\frac{AF}{FB}$ must give the same result.