Suppose $\Lambda$ is a bounded linear function on a Hilbert space $H$, given by an inner product with a unique fixed vector $h_0 \in H$ such that $\Lambda(h) = \langle h,h_0 \rangle$.
Set $M = \ker (\Lambda) := \{h \in H : \Lambda(h)=0\}$.
In a textbook proof, one can fix $z \in M^\perp$, scale $\Lambda(z)=1$, and then find that $$\Lambda(\Lambda(h)z-h)=0,$$ which means $\Lambda(h)z-h \in M$. This means furthermore that $$\langle \Lambda(h)z-h,z \rangle=0.$$ How does one then use the orthogonality of $\Lambda(z)-h$ and $z$ to deduce that $\Lambda(h)=\langle h,\frac z{\|z\|^2}\rangle$?
$$\langle \Lambda(h)z - h,z \rangle = \Lambda(h)\langle z,z\rangle -\langle h,z\rangle$$ This follows from the linearity (sesquilinearity in case of complex spaces)