Proving collinear points - geometry Straight line $p$ contains the orthocenter $H$ of an acute triangle $ABC$, where $p$ intersects sides $AB$ and $AC$ with $p$ intersecting $AC$ in the point $P$. Straight line $q$ also contains $H$, is perpendicular to $p$ and intersects $AB$ and $BC$ with $q$ intersecting BC in the point $Q$. Straight line containing $A$, parallel to $q$ and a straight line containing $B$, parallel to $p$ intersect in the point $R$. Prove that $P,Q$ and $R$ are collinear.
A question I came across in a practice book, I am not sure even where to start... I'm allowed to use trigonometry, similar triangles, point-line-plane.

 A: Let $B'$ be a point on $AC$ such that $\angle AB'B = 90°$. Denote the intersection point of $AR$ and $BB'$ by $B''$.
Case 1: When $B''$ is out of $\overline {BH}$

Joint $AH$, $B'N$ and $PB''$. Let $M$ be the intersection point of $BR$ and $QH$, and $N$ be the intersection point of $AR$ and $HP$.
Since $\triangle BMH$ ~ $\triangle HNB''$, $BM \times NB'' = MH \times HN$.
Also, $\angle AB'H = \angle ANH = 90°$ implies $B'B''NP$ and $AHNB'$ are cyclic quadrilateral.
$\Rightarrow \angle B''PN = \angle HB'N = \angle HAN = 90° - \angle AHN$
$\Rightarrow B''P \perp AH$, $B''P$//$BC$
$\Rightarrow \triangle B''PN$ ~ $\triangle QBM$
$\Rightarrow PN\times MQ = BM \times NB'' = MH \times HN = RN \times MR$
$\Rightarrow \triangle NPR$ ~ $\triangle MRQ$
$\Rightarrow \angle QRM = \angle RPN$
$\Rightarrow P$, $Q$ and $R$ are collinear
Case 2: When $B''$ lies on $\overline {BH}$
A → B and B → A
As same as Case 1
A: 
We draw the two altitudes $AX$ and $BY$ of the given scalene acute triangle $ABC$. For brevity, let $HP$ and $HQ$ equal $d$ and $e$ respectively. We also assume that $\measuredangle BCA=\omega$ and $\measuredangle HPA=\phi$.
We start our proof by doing some angle chasing in $\mathrm{Fig.\space 1}$. In the right angle triangle $AMP$,
$$\measuredangle PAM=90^o-\measuredangle MPA=90^o-\phi.$$
In a similar vein, by considering the right angle triangle $PYH$, we can state that,
$$\measuredangle YHP=90^o-\measuredangle HPY=90^o-\phi.$$
The line $HP$ is parallel to the line $BN$ and the altitude $BY$ is their transversal. Therefore, the two corresponding angles $\measuredangle YHP$ and $\measuredangle HBN$ are equal, i.e.
$$\measuredangle HBN=90^o-\phi.$$
As a consequence, in the right angle triangle $BNH$,
$$\measuredangle NHB=90^o-\measuredangle HBN=\phi. \tag{1}$$
Finally, consider the two right angle triangles $AXC$ and $CYB$, which share $\measuredangle BCA$. Since, we have assumed $\measuredangle BCA=\omega$,
$$\measuredangle CAX=\measuredangle YBC=90^o-\measuredangle BCA =90^o-\omega. \tag{2}$$
Now we have all the angles we need to proceed with the proof. First, consider the two triangles $AHP$ and $BQH$. They are similar triangles because of (1) and (2). Therefore, we shall write,
$$\frac{AP}{BH}=\frac{HP}{HQ}=\frac{d}{e}. \tag{3}$$
Let us pay our attention to the two right angle triangles $AMP$ and $BNH$. Because of (1), these two triangles are also similar. Hence, with the help of (3), we can state the following equation.
$$\frac{MP}{NH}=\frac{AP}{BH}=\frac{d}{e}. \tag{4}$$
Consider the right angle triangle $PMR$, where we denote $\measuredangle RPM$ as $\theta$, which then can be express as,
$$\theta = \tan^{-1}\left(\frac{MR}{PM}\right).$$
Since $HNRM$ is a rectangle, we have $MR=NH$. Therefore, using (4) we can write,
$$\theta = \tan^{-1}\left(\frac{NH}{PM}\right)= \tan^{-1}\left(\frac{e}{d}\right). \tag{5}$$
Now, visualize what happens if we join $P$ and $Q$. The triangle $PHQ$ becomes a right angle triangle. Therefore, using what we have assumed at the beginning of our proof $\left(\mathrm{i.e.}\space HP=d\space\space \mathrm{and}\space\space HQ=e \right)$, we can express $\measuredangle QPH$ as,
$$\measuredangle QPH = \tan^{-1}\left(\frac{HQ}{HP}\right)= \tan^{-1}\left(\frac{e}{d}\right) ,$$
which, according to (5), is equal to $\theta$. Therefore, $\measuredangle QPH = \measuredangle RPM$, which implies that the point $R$ lies on the line $PQ$.
