Proving a theorem in Simson Lines I was trying to answer this problem but I'm not really sure if what I'm doing is correct. What I'm arriving at is not conclusive and I believe I need help in solving this. 

"The simson line of diametrically opposite points on the circumcircle
  are perpendicular to each other and meet on the nine-point circle."

So far this has been my progress (cumulative efforts of the internet and my friends)


*

*Let $P$ lie on the circumcircle of triangle $ABC$

*Let $Q$ be the point diametrically opposite $P$

*$P_a$, $P_b$, $P_c$, and $Q_a$, $Q_b$, $Q_c$ are the feet perpendicular from $P$ and $Q$ onto sides $BC$, $AC$, and $AB$.

*$CP_aP$ and $CP_bP$ are right which makes the quadrilateral $CPP_aP_b$ cyclic*.The same thing applies to quadrilateral $QQ_bQ_aC$ because $QQ_bQ_a$ and $CQ_aQ_b$ are right angles. (Note: in the figure provided, I wasn't able to create the circles)

*Based on an Internet source, inscribed angles $CP_aP_b$ and $PP_aP_b$ are equal (this is a bit hazy because I don't know the reason why they are such)

*??? 

*??? 
*Note: Not sure what exact theorem states this but this is as far as I could remember: Given 2 right angles in a quadrilateral, then that quadrilateral is cyclic-- Would really be grateful if you could point out what theorem this is.

I would really appreciate if you can offer down a step by step proof (along with the theorem/property etc that supports the claim). Thank you very much!
PS. Will post the figure I drew soon. I think the mobile version of stackexchange doesn't support image uploading. Thank you!
 A: I couldn't write all the details, but I hope this will help. As you said, quadrilateral $CPP_aP_b$ is cyclic, that's because $\angle PP_aC=\angle CP_bP$ (it's not important for this step to say that they are right angles). Now I assume that you want to prove that 
$$ \angle CPP_b=\angle CP_aP_b$$
as it's written in the bottom of your figure. This is true just because these two angles are subtended by the same arc $CP_b$ of the circumcircle of $CPP_aP_b$. The same argument applies to the quadrilaterl $QCQ_aQ_b$ and we can deduce that
$$ \angle CQQ_a=\angle CQ_bQ_a.$$
Now comes the key step. We have that $\angle P_bPP_a=\angle P_bCP_a$, because they are subtended by the same arc $P_aP_b$; for the same reason $\angle Q_bQQ_a=\angle Q_bCQ_a$ because they are both subtended by $Q_aQ_b$. Now notice that $\angle P_bCP_a=\angle Q_bCQ_a$, this is obvious. So put all this together we have
$$\angle P_bPP_a=\angle P_bCP_a=\angle Q_bQQ_a=\angle Q_bCQ_a.$$
The angle $\angle QCP$ is a right angle, because by hypothesis $QP$ is a diameter of the circle, this implies that
$$\angle QCQ_b+\angle Q_bCQ_a+\angle P_aCP=\angle QCP=\pi/2 \tag{1} $$
Since $CQ_bQ$ is a right triangle we have that
$$\angle CQQ_a+\angle Q_bQQ_a+\angle QCQ_b=\pi/2$$ 
but we already know that $\angle Q_bQQ_a=\angle Q_bCQ_a$ , so 
$$\angle CQQ_a+\angle Q_bCQ_a+\angle QCQ_b=\pi/2 \tag{2}.$$
If we put together $(2)$ and $(1)$, we deduce
$$\angle P_aCP=\angle CQQ_a. $$ Repeating this argument we finally deduce that the two quadrilateral are similar, because they have the same internal angles. Hence $P_bP_a$ and $CP$ form the same angle as $QC$ and $Q_bQ_a$, this means that the two lines $P_bP_a$ ,$Q_bQ_a$ are just the lines $CP$,$QC$ rotated by the same angle, and since $CP \perp QC$, then also 
$$P_bP_a \perp Q_bQ_a.$$
