$ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ $\implies$ the triangle is equilateral? If in a triangle $ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ , then is the triangle equilateral ?

$\displaystyle a=2R\sin A\implies\cos A=R(\sin2A)$
$\displaystyle\implies\sum a\cos A=R(\sum\sin2A)$
Using Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle,  $\displaystyle\sum a\cos A=4R\prod\sin A=4R\prod2\sin\frac A2\cos\frac A2$
Again, $\displaystyle\frac{a+b+c}2=R(\sum \sin A)$
Using this,  $\displaystyle\frac{a+b+c}2=4R\prod\cos\frac A2$
As $\displaystyle0<A,B,C<\pi\implies\cos\frac A2$ (etc.) $\ne0,$ the problem reduces to $\displaystyle\prod2\sin\frac A2=1$
$\implies4\sin\frac A2\left(\cos\frac{B-C}2-\cos\frac{B+C}2\right)=1 $
As $\displaystyle B+C=\pi-A\implies \frac{B+C}2=\frac\pi2-\frac A2$
$\displaystyle\iff4\sin^2\frac A2-4\sin\frac A2\cos\frac{B-C}2+1=0 $ which is a Quadratic Equation in $\sin\frac A2$
As $A$ is real the discriminant  $\displaystyle\left(4\cos\frac{B-C}2\right)^2-4\cdot4=-16\sin^2\frac{B-C}2$ must be $\ge0$
$\displaystyle\implies\sin^2\frac{B-C}2=0\iff\cos(B-C)=1-2\sin^2\frac{B-C}2=1\implies B-C=2n\pi $
As $\displaystyle0<B,C<\pi\implies B=C\implies\sin\frac A2=\frac12$
Can you prove $A=60^\circ?$
