If three cevians are concurrent at a point and form triangles of equal area, the point is the centroid

Question

Let D,E,F be points on side BC,CA,AB of triangle ABC. The three cevians are concurrent at a point G. The areas of triangles BGD, CGE and AGF are equal. Prove that G is the centroid of ABC

I have constructed lines through D,E,F that are parallel to the sides but this didn't give me anything of use, I tried using Ceva but didn't learn anything useful and I constructed altitude from D,E,F to the cevians in hopes of the similar triangles giving me something useful. It seems most likely that proving that the areas of all the triangles are equal is the way to solve it but I can't see how to do this.

Answer 1

Let $\Delta$ = area BGD = area CGE = area AGF. Let $p=\frac{DC}{BD},q=\frac{EA}{CE},r=\frac{FB}{AF}$. By Ceva we have $pqr=1$.

So taking area BGD=1 the areas of the other small triangles are as shown.

Now AG/GD = area AGB/area DGB = $r+1$. But AG/GD = area AGC/area DGC = $\frac{1+q}{p}$, so $p(1+r)=1+q$. Similarly $q(1+p)=1+r,r(1+q)=1+p$. Adding we get $pq+qr+rp=3$. But by AM/GM $(pq+qr+rp)/3\ge(pqr)^{2/3}=1$ with equality iff $pq=qr=rp$. Hence $p=q=r=1$.

Hence D,E,F are the midpoints and so G is the centroid.

Answer 2

Reasoning by contradiction. If $G$ not centroid, then at least one of the points on the sides is not in the middle. Let $AE > EC$. By property chevian,product $(AE:EC)(CD:DB)(BF:FA)$ is unity. Hence, at least one of the last two relations less than 1. Let us consider two cases.

1) $CD < DB$. According to Van Aubel's Theorem, $CG:GF=CE:EA+CD:DB < 2$. Then area $S_{AGF}:S_{AGC}=GF:GC > \frac12$, while $S_{CGE}:S_{AGC}=CE:CA < \frac12$. So, ares $AGF$ and $CGE$ are not equal - a contradiction.

2) $BF < FA$. Here, by the same theorem $AG:GD=AE:EC+AF:FB > 2$. Then $S_{BGD}:S_{BGA}=GD:GA < \frac12$, and $S_{AGF}:S_{BGA}=AF:AB > \frac12$, and we arrive at a contradiction.