Prove that the intersection angle between the Simson lines of two triangles inscribed in the same circle it's the same for any point.

Prove that the intersection angle between the Simson lines of two triangles inscribed in the same circle it's the same for any point.
Suppose the triangles ABC and DEF share the circuncircle C, and P and Q are any diferent points on C. Let l, m be, the Simson lines of P related to ABC and DEF, and p, q the Simson lines of Q related to ABC and DEF, then i must prove the angle between l and m equals the angle between p and q.
I just have one Theorem about the Simson line:
Theorem: Let P, Q be two points on the circuncircle, C, of the triangle ABC. Let l, m be their respective Simson's lines. Then the angle between l an m equals to the half of the central angle POQ, where O is the center of C.
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Ismailatirf
Step 1
The theorem you cite tells you that the angle between l and p is $\frac{1}{2}\measuredangle POQ$. And the angle between m and q is the same. So you have
$\begin{array}{rl}\measuredangle lp& =\frac{1}{2}\measuredangle POQ\\ \measuredangle mq& =\frac{1}{2}\measuredangle POQ\end{array}$
Step 2
If you write ${\alpha }_{p}$ as the direction of p with respect to an arbitrary reference direction, and likewise for the other lines using the same reference, then you can also write this as
$\begin{array}{rl}{\alpha }_{p}& ={\alpha }_{l}+\frac{1}{2}\measuredangle POQ\\ {\alpha }_{q}& ={\alpha }_{m}+\frac{1}{2}\measuredangle POQ\end{array}$
Step 2
Now subtract both sides to get $\measuredangle pq={\alpha }_{q}-{\alpha }_{p}={\alpha }_{m}-{\alpha }_{l}=\measuredangle lm$