I've send this question a little while ago now I edit it
Let $A$ be a Banach algebra with identity $e_A$
prove that $e_A$ is an extreme point of $B_A = \{a\in A: \|a\| \leq 1\} $
Recall that $x$ is an extreme point of $B_A$ if $x = \frac{1}{2}(x_1+x_2)$ for some $x_1,x_2\in B_A$ implies that $x= x_1=x_2.$
My attempt: $x,y \in B_A$, $t\in (0,1)$ and $e_A=tx+(1-t) y$
it is easy to see that $$\|x\|=\|y\|=1$$ therefore $$\|e_A-tx\|=\|(1-t) y\|<1 {\text{and }}\|e_A(1-t) y\|=\|tx \|<1$$
this implies that $$tx, (1-t) y \in Inv A$$ where $Inv A$ is the set of all invertible elements of $A$
I can't continue from here