With $P_A=A(A'A)^{-1}A'$ and $M_A=I-P_A$ denoting the usual orthogonal projection matrices, I'm trying to find an alternative proof to $$ P_{[X,Z]}=P_X+M_XZ(Z'M_XZ)^{-1}Z'M_X.\tag{i} $$ I already proved (i) by writing $$ P_{[X,Z]}=(P_X+M_X)P_{[X,Z]}=P_XP_{[X,Z]}+M_XP_{[X,Z]}\\ =P_X+[0,Z]\begin{pmatrix}X'X&X'Z\\Z'X&Z'Z\end{pmatrix}^{-1}\begin{pmatrix}X'\\Z'\end{pmatrix}. $$ Then I simplified the inversion above using the very last formula given here. From that point on, some algebra simplifications produced (i).
This approach is a bit tedious and it relies on a block inversion formula that I will never be able to memorize. Does anyone know a more elegant approach please?
Note that the range of $[X,Z]$ is the same as the range of $[X,M_XZ]$ where from the second block we only removed the components which are already in the range of $X$ and in addition, we made these vectors orthogonal to the range of $X$. So (since $M_XX=0$) $$ \begin{split} P_{[X,Z]}&=P_{[X,M_XZ]}=[X,M_XZ]\left(\begin{bmatrix}X'\\Z'M_X\end{bmatrix}[X,M_XZ]\right)^{-1}\begin{bmatrix}X'\\Z'M_X\end{bmatrix} \\&=[X,M_XZ]\left(\begin{bmatrix}X'X&0\\0&Z'M_XZ\end{bmatrix}\right)^{-1}\begin{bmatrix}X'\\Z'M_X\end{bmatrix}=P_X+M_XZ(Z'M_XZ)^{-1}Z'M_X. \end{split} $$ You won't completely avoid inverting a block matrix, which is, however, block diagonal now.