Show that i is idempotent

Question

Let $R$ be a (non commutative) Ring with unity, $I$, $K$ Left-Ideals of $R$ such that the sum $I \oplus K$ is direct, with $R = I \oplus K$ and $1 = i + k$, $i \in I$, $k \in K$. I have to prove that $i^{2} = i$. I only know that $1^{2} = 1 = i + k = (i + k)^{2} = i^{2} + ki + ik + k^{2} \Rightarrow i = i^{2} + ki$. But I have shown that $ik = -ik$. But I don't know how to the next step is.

Answer 1

$i=i(i+k)=i^2+ik $ and $i=(i+k)i =i^2+ki$, so $ik=ki\in I\cap K=\{0\}$.

Therefore $i=i^2$ from the first equation.