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.

$$a^2+b^2+c^2=ac+ab\sqrt3$$ The above equation can be re-written as
$$\frac{a^2}{4}-ac+c^2+\frac{3a^2}{4}-ab\sqrt3+b^2=0$$
which is
$$(\frac{a}{2}-c)^2+(\frac{\sqrt3 a}{2}-b)^2 = 0$$
which implies that
$\frac{a}{2} = c$ and $\frac{\sqrt3 a}{2}= b$
.Based on this it can be concluded that the ratio of sides is 1:$\sqrt{3}$:2, which is a right angled-triangle.I hope this was helpful.
