# 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.

• Welcome to MSE! What have you tried to make headway in the problem so far? – Prakhar Nagpal Jan 10 '18 at 15:24
• This problem is really difficult for me and I have not found a way to solve it. – Mary Jan 10 '18 at 15:34
• And are you able to follow the given solutions? – Prakhar Nagpal Jan 10 '18 at 15:50

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.
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!
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.