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.
Since $0=a^2+b^2+c^2-ac-ab\sqrt{3}=\left(b-\frac{\sqrt{3}}2a\right)^2+\left(c-\frac12a\right)^2\,,$ we have $b=\frac{\sqrt{3}}2a$ and $c=\frac{1}{2}a$.  Thus, $\angle A=\frac{\pi}{2}$, $\angle B=\frac{\pi}{3}$, and $\angle C=\frac{\pi}{6}$.

It can be shown that, if $\alpha,\beta,\gamma\in(0,\pi)$ and $\alpha+\beta+\gamma=\pi$, then there is a unique triangle $ABC$, up to scaling, with $BC=a$, $CA=b$, and $AB=c$ such that $$a^2+b^2+c^2=2bc\cos(\alpha)+2ca\cos(\beta)+2ab\cos(\gamma)\,.$$
To show this, one observes that the matrix $$\textbf{X}:=\begin{bmatrix}1&amp;-\cos(\gamma)&amp;-\cos(\beta)\\-\cos(\gamma)&amp;1&amp;-\cos(\alpha)\\-\cos(\beta)&amp;-\cos(\alpha)&amp;1\end{bmatrix}$$ is positive-semidefinite and the eigenspace associated with the eigenvalue $0$ of $\textbf{X}$ is spanned by $\begin{bmatrix}\sin(\alpha)\\\sin(\beta)\\\sin(\gamma)\end{bmatrix}$.
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 $b=c$.