<s>[INST] 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.

 [/INST] 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!
</s>