If in $a\mathrm{△}ABC,a^2+b^2+c^2=ac+\sqrt{3}$ ab then the triangle 

$\begin{array}{l}{a}^2+b^2+c^2=ac+\sqrt{3}ab\\ \frac{a^2}{4}+c^2-ac+\frac{3a^2}{4}+b^2-\sqrt{3}ab=0\\ {\left(\frac{a}{2}-c\right)}^2+{\left(\frac{\sqrt{3}a}{2}-b\right)}^2=0\end{array}$

$\begin{array}{l}\frac{a}{2}-c=0,\frac{\sqrt{3}a}{2}-b=0\\ a=2c,\sqrt{3}a=2b\\ a=\frac{2b}{\sqrt{3}}=2c=\lambda \Rightarrow b=\frac{\sqrt{3}}{2}\lambda ,c=\frac{\lambda }{2}\end{array}$

${\mathrm{b}}^2+{\mathrm{c}}^2={\mathrm{a}}^2\text{, hence the }\mathrm{\Delta }$ is right angle but not isoceless