In a triangle \[ABC\], if \[b + c = 2a\] and \[\angle A = {60^{\circ}}\], then the \[\Delta ABC\] is
A. Scalene
B. Equilateral
C. Isosceles
D. Right angled

Hint:
 In the given question, we need to find the type of \[\Delta ABC\]. For this, we need to use projection formula. After simplification of this, we need to determine the angles of a triangle. Based on this, we can find the type of triangle.



Formula Used:
The following formula is used for solving the given question.
Suppose \[a,b\] and \[c\] are the sides of a triangle \[ABC\] and also \[A,B\] and \[C\] are the angles of a triangle \[ABC\] then the projection rule is \[b = a\cos C + c\cos A\] and \[c = a\cos B + b\cos A\].
Also, \[\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \]
\[\cos \left( A \right) + \cos (B) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
\[\angle A + \angle B + \angle C = {180^{\circ}}\]



Complete step-by-step answer:
Assume that \[a,b\] and \[c\] are the sides and \[A,B\] and \[C\] are the angles of a triangle \[ABC\].
Given that, \[b + c = 2a\] and \[\angle A = {60^{\circ}}\].
By using the projection formula, we get
\[b = a\cos C + c\cos A.....(1)\]
\[c = a\cos B + b\cos A.....(2)\]
By adding the above two equations, we get
\[b + c = a\{ \cos C + \cos B\} + (b + c)\cos A\]
\[b + c = a\{ 2\cos \left( {\dfrac{{C + B}}{2}} \right)\cos \left( {\dfrac{{C - B}}{2}} \right)\} + (2a)\cos {60^{\circ}}\]
But \[\dfrac{A}{2} + \dfrac{B}{2} + \dfrac{C}{2} = \dfrac{\pi }{2} \]
\[\Rightarrow \dfrac{B}{2} + \dfrac{C}{2} = \dfrac{\pi }{2} - \dfrac{A}{2}\]
So, we get
\[b + c = a\{ 2\cos \left( {\dfrac{\pi }{2} - \dfrac{A}{2}} \right)\cos \left( {\dfrac{{C - B}}{2}} \right)\} + (2a)\dfrac{1}{2}\]
By simplifying, we get
\[b + c = a\{ 2\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{{C - B}}{2}} \right)\} + a\]
Now, put \[b + c = 2a\] in the above equation.
\[2a = a\{ 2\sin \left( {{{30}^{\circ}}} \right)\cos \left( {\dfrac{{C - B}}{2}} \right)\} + a\]
\[a = a\{ 2\left( 1 \right)\cos \left( {\dfrac{{C - B}}{2}} \right)\} \]
By simplifying further, we get
\[1 = 1 \times \cos \left( {\dfrac{{C - B}}{2}} \right)\]
Thus, \[\cos {0^{\circ}} = \cos \left( {\dfrac{{C - B}}{2}} \right)\]
By comparing, we get
\[\dfrac{{C - B}}{2} = 0\]
By solving, we get
\[C = B\]
Also, \[\angle A + \angle B + \angle C = {180^{\circ}}\]
But \[\angle A = {60^{\circ}}\]and \[C = B\]
Hence, \[{60^{\circ}} + \angle B + \angle B = {180^{\circ}}\]
\[2\angle B = {180^{\circ}} - {60^{\circ}}\]
This gives \[2\angle B = {120^{\circ}}\]
So, \[\angle B = {60^{\circ}}\]
Thus, \[\angle C = {60^{\circ}}\]
Here, we can say that \[\angle A = \angle B = \angle C = {60^{\circ}}\]
Thus, the given triangle is an equilateral triangle.
Therefore, the correct option is (B).



Note:
Many students make mistakes in the calculation part as well as in writing projection formulae. This is the only way, through which we can solve the example in the simplest way. Also, it is essential to use proper trigonometric identities and their formulae to get the desired result.