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.

 A: Let's connect G and H. $GH$ is parallel to $AB$ (midsection theorem). As Michael Rozenberg noted, ABGH is a cyclic trapezoid, which is only possible when trapezoid is isosceles. Hence, $AG=BH$, $AC=BC$. Now from $\triangle BCG$ we see that the side opposite 30 degree angle ($CG$) is half of the other side ($CB$) which means $\angle CGB$ is a right angle and $\angle ACB$ is 60 degrees. Done.  
A: Since $\measuredangle GBH=\measuredangle GAH$, we see that $ABHG$ is cyclic.
Thus, by Ptolemy $$AB\cdot GH+BH\cdot AG=AH\cdot BG$$ or in the standard notation
$$\frac{c^2}{2}+\frac{ab}{4}=\frac{1}{4}\sqrt{(2b^2+2c^2-a^2)(2a^2+2c^2-b^2)},$$ 
which after squaring of the both sides gives
$$(a-b)^2(a+b-c)(a+b+c)=0$$ or
$$a=b.$$
Now, by law of cosines for $|delta AHC$ we obtain
$$\cos30^{\circ}=\frac{AH^2+AC^2-HC^2}{2AH\cdot AC}$$ or
$$\frac{\sqrt3}{2}=\frac{\frac{1}{4}(2b^2+2c^2-a^2)+b^2-\frac{a^2}{4}}{2\cdot\frac{1}{2}\sqrt{2b^2+2c^2-a^2}b}$$ or
$$\sqrt3b\sqrt{2b^2+2c^2-a^2}=3b^2+c^2-a^2$$ or
$$3b^2(a^2-b^2)+(a^2-c^2)^2=0,$$ which gives $$a=c$$ and we are done!
A: 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.
