Show that the triangle is equilateral triangle If a triangle $ABC$ has the equality $$h_a\cdot\sqrt{3} +\frac{a}{2}= b + c$$ $h_a$ is the height from $A$, then show that the triangle is equilateral.
Using sine rule, I tried to show that bring equality to a form of showing that $A=B=C=\frac{\pi}{3}$ but I managed. 
Does anyone have a solution?
Thank you very much! 
 A: Considering the area of the triangle, we have
$$\frac 12ah_a=\frac 12bc\sin\angle{BAC},$$
i.e.
$$\sin\angle{BAC}=\frac{ah_a}{bc}=\frac{a(2b+2c-a)}{2\sqrt 3\ bc}\tag1$$
Using $(1)$ and the law of cosines,
$$\left(\frac{b^2+c^2-a^2}{2bc}\right)^2=\cos^2\angle{BAC}=1-\left(\frac{a(2b+2c-a)}{2\sqrt 3\ bc}\right)^2,$$
i.e.
$$12b^2c^2-a^2(2b+2c-a)^2-3(b^2+c^2-a^2)^2=0$$
Expanding the LHS,
$$4a^4-4a^3b-4a^3c-2a^2b^2+8a^2bc-2a^2c^2+3b^4-6b^2c^2+3c^4=0$$
Making two groups,
$$a^2(4a^2-4ab-4ac-2b^2+8bc-2c^2)+3(b^2-c^2)^2=0$$
then
$$a^2(4a^2-4ab-4ac+b^2-3b^2+2bc+6bc+c^2-3c^2)+3(b^2-c^2)^2=0$$
and rearranging
$$a^2(4a^2+b^2+c^2-4ab-4ac+2bc-3b^2+6bc-3c^2)+3(b^2-c^2)^2=0$$
to have
$$a^2((2a-b-c)^2-3(b-c)^2)+3(b^2-c^2)^2=0$$
then 
$$a^2(2a-b-c)^2=3a^2(b-c)^2-3(b^2-c^2)^2$$
$$a^2(2a-b-c)^2=3(b-c)^2(a^2-(b+c)^2)$$
$$a^2(2a-b-c)^2=3(b-c)^2(a-b-c)(a+b+c)$$
Now, the LHS is non-negative, and the RHS is non-positive as $a-b-c\lt 0$, from which we have to have
$$a^2(2a-b-c)^2=3(b-c)^2(a-b-c)(a+b+c)=0,$$
i.e.
$$2a-b-c=b-c=0\quad\Rightarrow\quad a=b=c$$
as desired.
