$ 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 ?

This  is a 1990 Russian maths contest problem
$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, so 
$$a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $$
$$\Longleftrightarrow$$
$$a^4+b^4+c^4-2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)=0\tag{1}$$
There are several ways to show (1)
In fact
$$a^4+b^4+c^4-2(a^2b^2+b^2c^2+c^2a^2)+abc(a+b+c)$$
$$=(a+b+c)\Big(abc-(-a+b+c)(a-b+c)(a+b-c)\Big)$$
It is easy to show that
$$abc-(-a+b+c)(a-b+c)(a+b-c)=0\Longleftrightarrow a=b=c$$
another method to prove (1)
one can use Schur's inequality
$$\sum a^2(a-b)(a-c)\geq0$$
