<s>[INST] In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these

The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+\sin^2B+\sin^2C=\sin A\sin C+\sin A\sin B\sqrt3$

I am stuck here.
 [/INST] If $\triangle ABC$ is equilateral, then $3x^2=x^2(\sqrt{3}+1)$, absurd.
Now, note that the condition is symmetric respect to $b$ and $c$, ie, if $(a,b,c)$ satisfies, $(a,c,b)$ too, and then $\triangle ABC$ is isosceles with $b=c$.
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