For triangle $ABC$ there are median lines $AH$ and $BG$ with $\angle CAH=\angle CBG={{30}^{0}}$ . Prove that $ABC$ is the equilateral triangle. For triangle $ABC$ there are  median lines $AH$ and $BG$ with $\angle CAH=\angle CBG={{30}^{0}}$. Prove that $ABC$ is the equilateral triangle.


With the standard notation, if $M_A$ is the midpoint of $BC$,
$$ AM_A^2 = \frac{2b^2+2c^2-a^2}{4},\quad AC^2 = b^2,\quad CM_A^2=\frac{a^2}{4} $$
hence if $\widehat{M_A A C}=30^\circ$ we have
$$ \cos 30^\circ = \frac{\sqrt{3}}{2} = \frac{AM_A^2+AC^2-CM_A^2}{2\cdot AM_A\cdot AC}=\frac{-a^2+3b^2+c^2}{4b\cdot AM_A} $$
and
$$ 2b\sqrt{3}\cdot AM_A = -a^2+3b^2+c^2 $$
$$ 2a\sqrt{3}\cdot BM_B = -b^2+3a^2+c^2 $$
lead to
$$ 3b^2(2b^2+2c^2-a^2)=(-a^2+3b^2+c^2)^2 $$
$$ 3a^2(2a^2+2c^2-b^2)=(-b^2+3a^2+c^2)^2 $$
or
$$ a^4+3b^4+c^4 = 3a^2 b^2+2a^2 c^2 $$
$$ b^4+3a^4+c^4 = 3a^2 b^2+2b^2 c^2 $$
so by difference $b^4-a^4 = c^2(a^2-b^2)$. If we assume $a\neq b$ we get $a^2+b^2+c^2=0$, which is absurd, hence $a=b$ and
$$ a^4+c^4 = 2a^2 c^2 $$
leads to $a=c$, too.
