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(B) Given $b + c = 2a$ .....$(i)$Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$Since $\angle A = 60^\circ$,we have $\frac{1}{2} = \f

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Equilateral

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(B) Given $b + c = 2a$ .....$(i)$Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$Since $\angle A = 60^\circ$,we have $\frac{1}{2} = \frac{b^2 + c^2 - a^2}{2bc}$$bc = b^2 + c^2 - a^2$$a^2 = b^2 + c^2 - bc$From $(i)$,$a = \frac{b+c}{2}$,so $a^2 = \frac{(b+c)^2}{4} = \frac{b^2 + c^2 + 2bc}{4}$Equating the two expressions for $a^2$: $\frac{b^2 + c^2 + 2bc}{4} = b^2 + c^2 - bc$$b^2 + c^2 + 2bc = 4b^2 + 4c^2 - 4bc$$3b^2 + 3c^2 - 6bc = 0$$3(b^2 + c^2 - 2bc) = 0$$3(b - c)^2 = 0$Thus,$b = c$.Substituting $b = c$ into $(i)$,$2b = 2a \implies b = a$.Since $a = b = c$,the triangle is equilateral.

In a triangle $ABC$,if $b + c = 2a$ and $\angle A = 60^\circ$,then $\Delta ABC$ is

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