Prove that idempotent operator $E$ is self-adjoint if and only if $EE^∗$ = $E^∗E$

Suppose $E^2=E$. Then $E$ has a basis of eigenvectors with eigenvalues $0$ and $1$ because every vector can be written as $$ x = Ex + (I-E)x, $$ and $\;$ $Ex$, $(I-E)x$ $\;$ satisfies \begin{align} E(Ex)&=1\cdot Ex,\\ E(I-E)x &= 0 \cdot (I-E)x. \end{align} $E$ is selfadjoint iff these eigenspaces are mutually orthogonal, which is equivalent to the condition that $$ \langle Ex,(I-E)y\rangle=0,\;\;\; \forall x,y, \\ \iff \langle x,E^*(I-E)y\rangle = 0, \;\;\; \forall x,y, \\ \iff E^*(I-E)y=0,\;\; \forall y \\ \iff E^*(I-E) = 0 \\ \iff E^* = E^*E $$ The last condition holds iff $E^*=E^*E = (E^*E)^*=E$.