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: let $x,y \in B_A$, $t\in (0,1)$ 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
Can I get some hints?
Suppose $1=\frac{1}{2}x_1 +\frac{1}{2}x_2$ with $x_1,x_2\in B_A$ where $1$ denotes the unit of $A$. Let $B_{A^*}$ denote the unit ball in $A^*$ and consider the set $Y=\{\psi \in B_{A^*}: \psi(1-x_1)=||1-x_1||\}$. It is convex and compact in the weak*-topology so by Krein-Milman $Y$ has an extremal point $\phi$. $\phi$ must be extremal even in $B_{A^*}$: if $\phi=\frac{1}{2}\phi_1+\frac{1}{2}\phi_2$ for $\phi_1, \phi_2 \in B_{A^*}$, the equation $||1-x_1||=\phi(1-x_1)=\frac{1}{2}\phi_1(1-x_1)+\frac{1}{2}\phi_2(1-x_1)$ along with $|\phi_1(1-x_1)|, |\phi_2(1-x_1)|\leq ||1-x_1||$ already implies that $\phi_1(1-x_1)=\phi_2(1-x_1)=||1-x_1||$ so $\phi_1, \phi_2\in Y$, which by extremality in $Y$ implies that $\phi_1=\phi_2=\phi$.
Now consider the decomposition $\phi=\frac{1}{2}\phi(\cdot x_1)+\frac{1}{2}\phi(\cdot x_2)$. Clearly $\phi(\cdot x_1),\phi(\cdot x_2)\in B_{X^*}$ so by extremality $\phi=\phi(\cdot x_1)$. In particular $\phi(1)=\phi(x_1) \rightarrow ||1-x_1||=\phi(1-x_1)=0$. Hence $x_1=x_2=1$.