Orthogonal Block Matrix Given $A \in \mathbb{R^{\text{nxn}}}$, $B \in \mathbb{R^{n\text{x}m}}$ and $C \in \mathbb{R^{m\text{x}m}}$ such that
$$ M = \begin{bmatrix}
    A & B \\
    0 & C 
\end{bmatrix} \in \mathbb{R^{(n+m)\text{x}(n+m)}} $$
a block Matrix.
Prove that: $M$ is orthogonal $\iff$ $A$ and $C$ are orthogonal and $B = 0$
My solution idea:
$\Longleftarrow$
Since $A$ is orthogonal, its columns form a linearly independent orthonormal set of vectors. It follows that the first $n$ columns of $M$ are orthonormal, because a vector $v_{i}$ of the form $(a_{1,i},a_{2,i},...a_{n,i},0,..,0)$ for $i = 1,..n$ still has Norm equal to $1$ and $\langle\ v_{i},v_{j} \rangle = 0$ for $j = 1,..,n$ given $i \neq j$
Same logic applies to columns vectors of $M$ from $n+1$ to $m$ since $C$ is orthogonal. So, taken together $\{v_{1},...,v_{n},v_{n+1},...,v_{m}\}$ they build an orthonormal Basis for $\mathbb{R^{(n+m)}}$ and it follows that $M$ is orthogonal.
$\Longrightarrow$
Reversing the argument:
Since $M$ is an orthogonal Matrix, its columns vectors build an orthonormal Basis for $\mathbb{R^{(n+m)}}$
Given that $a_{i,j} = 0$ for $i = m,..,n+m$ and $j =1,..,n$ It follows that the columns vectors of $A$ build an orthonormal Basis for $\mathbb{R^{n}}$ and $A$ is orthogonal.
Now to prove $B = 0$ confuses me. Because the columns vectors of $M$ are orthonormal, it has to be that the inner product of each column vector of $A$ with the columns vectors of $B$ is zero. Is this enough to follow that $B= 0$?
If $B = 0$ then it's easy to follow that $C$ is orthogonal.
Any help/comments would be great!

The $n+i$-th column of $M$ must be orthogonal to all the first $n$ columns of $M$. If you write the dot product, it's clear that the dot product of the $n+i$-th column of $M$ and the $k$-th column of $M$ (for $k\leq n$) is equal to the dot product of the $i$-th column of $B$ and the $k$-th column of $A$.
Therefore, the above demand is equivalent to demanding that the $i$-th column of $B$ must be orthogonal to all the $n$ columns of $A$. 
Since the columns of $A$ span $\mathbb R^n$ (orthogonal nonzero vectors are always linearly independent), the only vector orthogonal to all the columns of $A$ is the zero vector.
