In Delta ABC, if {a^2} + {b^2} + {c^2} = ac + | vedclass

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(C) Given,${a^2} + {b^2} + {c^2} = ac + ab\sqrt{3}$Rearranging the terms,we get: ${a^2} + {b^2} + {c^2} - ac - ab\sqrt{3} = 0$We can rewrite this as:

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(C) Given,${a^2} + {b^2} + {c^2} = ac + ab\sqrt{3}$Rearranging the terms,we get: ${a^2} + {b^2} + {c^2} - ac - ab\sqrt{3} = 0$We can rewrite this as: $\left(\frac{a^2}{4} - ac + c^2\right) + \left(\frac{3a^2}{4} - ab\sqrt{3} + b^2\right) = 0$This simplifies to: ${\left(\frac{a}{2} - c\right)^2} + {\left(\frac{a\sqrt{3}}{2} - b\right)^2} = 0$For the sum of squares to be zero,each term must be zero:$\frac{a}{2} - c = 0 \implies a = 2c$$\frac{a\sqrt{3}}{2} - b = 0 \implies b = \frac{a\sqrt{3}}{2}$Now,check the relation between sides: ${b^2} + {c^2} = {\left(\frac{a\sqrt{3}}{2}\right)^2} + {\left(\frac{a}{2}\right)^2} = \frac{3a^2}{4} + \frac{a^2}{4} = \frac{4a^2}{4} = {a^2}$Since ${b^2} + {c^2} = {a^2},$ the triangle is right-angled.

In $\Delta ABC,$ if ${a^2} + {b^2} + {c^2} = ac + ab\sqrt{3},$ then the triangle is:

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