Problem: "For $f,g$ in an inner product space, $g\neq0$, the projection of $f$ on $g$ is the vector, $\frac{\langle f,\,g\rangle}{\|g\|^2}g$. Show that the two vectors $\frac{\langle f,\,g\rangle}{\|g\|^2}g$ and $f-\frac{\langle f,\,g\rangle}{\|g\|^2}g$ are orthogonal."
Attempt: $$\langle\frac{\langle f,\,g\rangle}{\|g\|^2}g,f-\frac{\langle f,\,g\rangle}{\|g\|^2}g\rangle=\langle\frac{\langle f,\,g\rangle}{\|g\|^2}g,f\rangle-\langle\frac{\langle f,\,g\rangle}{\|g\|^2}g,\frac{\langle f,\,g\rangle}{\|g\|^2}g\rangle$$
$$=\frac{\langle f,\,g\rangle}{\|g\|^2}\langle g,f\rangle-\left|\frac{\langle f,\,g\rangle}{\|g\|^2}\right|^2\langle g,g\rangle$$
$$=\frac{\langle f,\,g\rangle}{\|g\|^2}\left(\langle g,f\rangle-\frac{\langle f,\,g\rangle}{\|g\|^2}\langle g,g\rangle\right)$$
I'm not too sure where to go from here. I have a feeling that I should be making use of a geometric fact to do with how we have constructed $\frac{\langle f,\,g\rangle}{\|g\|^2}g$ and $f-\frac{\langle f,\,g\rangle}{\|g\|^2}g$. What am I missing here?
Now, just use that fact that $\langle f,g\rangle=\langle g,f\rangle$ and that $\langle g,g\rangle=\|g\|^2$.