If $e$ and $f$ are idempotent and $e(e+f)f=e+f$, then prove that $e=f$

Question

Let $R$ be an associative ring and let the idempotents $e$ and $f$ belongs to $R$. Then prove that

$$e(e+f)f=e+f \iff e=f$$

The only if part is very easy and I have proved what about the $(\implies)$ part.

Answer 1

Proving the $\implies$ part

\begin{eqnarray} e(e+f)f &=& e+f \\ (e+ef)f &=& e+f \\ ef+ef &=& e+f \\ \end{eqnarray}

Now multiply above equation by $e$ (from left) and $f$ (from right) separately. We get

$$ ef+ef = e+ef $$

and,

$$ ef+ef = ef+f $$

Therefore,

$$ e+ef = ef + f $$

which implies $e=f$ (since addition is commutative)