If both roots of the Quadratic Equation are similar then prove that If both roots of the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that $2a=b+c$.

Things should be known:
  
  
*
  
*Roots of a Quadratic Equations can be identified by:
The roots can be figured out by: 
  $$\frac{-b \pm \sqrt{d}}{2a},$$ 
  where
  $$d=b^2-4ac.$$
  
*When the equation has equal roots, then $d=b^2-4ac=0$.
  
*That means $d=(b-c)^2-4(a-b)(c-a)=0$


As the two roots are equal the discriminant must be equal to $0$.
$$(b-c)^2-4(a-b)(c-a)=(a-b+c-a)^2-4(a-b)(c-a)=\{a-b-(c-a)\}^2=(2a-b-c)^2=0 \iff 2a-b-c=0$$

Alternatively, solving for $x,$ we get $$x=\frac{-(b-c)\pm\sqrt{(b-c)^2-4(a-b)(c-a)}}{2(a-b)}=\frac{c-b\pm(2a-b-c)}{2(a-b)}=\frac{c-a}{a-b}, 1$$ as $a-b\ne 0$ as $a=b$ would make the equation linear. 
So, $$\frac{c-a}{a-b}=1\implies c-a=a-b\implies 2a=b+c$$
