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 $d=(b-c)^2-4(a-b)(c-a)=0$


Edit: $(b-c)^2=4(a-b)(c-a)$, so $a-b$ and $c-a$ are either both nonnegative or both nonpositive. In the former case, put $x=a-b$ and $y=c-a$; in the latter case, put $x=-(a-b)$ and $y=-(c-a)$. Now the A.M. $\ge$ G.M. inequality says that
$$
\frac{x+y}2\ge\sqrt{xy},\ \text{ i.e. } (x+y)^2\ge4xy
$$
for $x,y\ge0$, and equality holds if and only if $x=y$. 
