How show that $ABC$ is equilateral? Let $D$, $E$ and $F$ be three points on sides $BC$,$AC$ and $AB$ of triangle $ABC$ such that lines $AD$, $BE$ and $CF$ concur at point $M$. If three trianles $MDB$, $MCE$ and $MAF$ have equal areas and equal perimeters. How show that $ABC$ is equilateral?
 A: This problem gave me a really hard time, lots of steps are involved.

Step 1. If $ABC$ is equilateral and $MAF,MBD,MCE$ have the same area, then $M$ is the center of $ABC$.

Given that $[x,y,z]$ are the trilinear coordinates of $M$, the areas of our triangles are proportional to: $$\frac{xy}{x+z},\frac{yz}{x+y},\frac{xz}{y+z}$$
hence $M=[1,1,1]$ is the only chance.

Step 2. If the areas of $MAF,MBD,MCE$ are equal, then $M$ is the centroid of $ABC$.

Take an affine map $\Phi$ that brings our original triangle into an equilateral one. Since the affine maps preserve the ratio between the areas, $\Phi(M)$ is the centroid of $\Phi(ABC)$ due to Step1, hence $M$ is the centroid of $ABC$ and $D,E,F$ are the midpoints of the sides.
Call now $m_a,m_b,m_c$ the lengths of the medians through $A,B,C$ and $a,b,c$ the lengths of the sides $BC,AC,AB$ respectively.

Step 3. If $a>b$, then $m_a<m_b$.

This just follows from the Stewart's theorem, granting:
$$ 4m_a^2 = 2b^2+2c^2-a^2,\qquad 4m_b^2=2a^2+2c^2-b^2.$$

Step 4. If $a>b$, then $\frac{a}{2}+\frac{m_a}{3}>\frac{b}{2}+\frac{m_b}{3}$.

This is the crucial part. It is straightforward to check that the inequality is equivalent to:
$$ a^2+b^2-ab > 2c^2-4m_a m_b,$$
but since $a^2+b^2\geq 2ab$, it is sufficient to prove the weaker inequality:
$$2c^2-ab< 4m_a m_b$$
that is equivalent to:
$$a^4+b^4 < 2a^2b^2 + a^2c^2+b^2 c^2+2abc^2$$
or to:
$$(a^2-b^2)^2 < (a+b)^2 c^2$$
or to:
$$(a-b)^2< c^2$$
that is just the triangle inequality $|a-b|<c$. Phew.
Suppose now that $ABC$ is not equilateral. We can assume without loss of generality that $a\geq b\geq c$ and at least one of the two inequalities holds tight. But due to Step3 and Step4 we have:
$$\left(\frac{a}{2}+\frac{m_a}{3}\right)+\frac{2m_b}{3}>\left(\frac{c}{2}+\frac{m_c}{3}\right)+\frac{2m_a}{3},$$
hence the triangles $MAF,MBD$ and $CEF$ cannot all have the same perimeter, contradiction.
