Proving independence between linearity and quadratic forms using independence of linear and quadratic form

Question

I'm reading through these lecture notes online http://www.pitt.edu/~wahed/teaching/2083/fall09/Lecture309.pdf

And on page 103 he notes the following theorem

If $X \sim N(\mu, \Sigma)$ and $A^T = A$, $B^T = B$, then $X^T A X \perp BX$ if and only if $B \Sigma A = 0$. (The assumption of symmetry is listed earlier in the text)

The proof is ommited unfortunately, but I want to understand why this is the case. In particular, if we assume that

1. $X^T A X \perp X^TBX \iff B \Sigma A = 0$
2. $ A X \perp BX \iff B \Sigma A = 0$

How can we prove the above theorem. In other words, if linear forms are independent of one another, and quadratic forms are independent of one another (given $B \Sigma A = 0$), can you then prove that linear forms are independent of quadratic forms?

Or is that not sufficient to prove the above theorem?

---

My attempt in one direction (incomplete) is to note that if $A$ is symmetric it can be decomposed into $A = QDQ^T$ where $D$ is diagonal and $Q$ orthogonal. Hence,

Let $B \Sigma A = 0$

Then by independence of linear forms we have $BX \perp AX$

$ \therefore BX \perp QDQ^T X$

$ \therefore BX \perp (QDQ^TX)^T QDQ^T X$

$\therefore BX \perp X^T QDQ^T QDQ^T X$

$\therefore BX \perp X^T QDDQ^T X$

Since $Q$ is orthogonal. But here I'm stuck, since $D$ is diagonal, so $D \neq D^2$ and so $QDDQ^T \neq A$.

As for the other direction, I have no idea where to begin.

Thoughts?