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August 17th, 2018, 04:08 AM  #1 
Member Joined: Jan 2018 From: Belgrade Posts: 50 Thanks: 2  Prove that P, Q and R are collinear
Straight line $\displaystyle p$ contains orthocenter $\displaystyle H$ of the acute triangle $\displaystyle ABC$ and intersects its edges $\displaystyle AB$ and $\displaystyle CA$, and edge $\displaystyle CA$ at point $\displaystyle P$. Straight line $\displaystyle q$ perpendicular to $\displaystyle p$ contains point $\displaystyle H$ also, intersects edges $\displaystyle AB$ and $\displaystyle BC$, and edge $\displaystyle BC$ at point $\displaystyle Q$. Straight line through $\displaystyle A$ parallel to $\displaystyle q$, and staight line through $\displaystyle B$ parallel to $\displaystyle p$ intersect at point $\displaystyle R$. Prove that points $\displaystyle P$, $\displaystyle Q$ and $\displaystyle R$ are collinear.

September 14th, 2018, 12:34 PM  #2 
Member Joined: Jan 2018 From: Belgrade Posts: 50 Thanks: 2 
I tried to make some progress. See the picture. We can write: $\displaystyle DH=RE$, and $\displaystyle RD=HE$. $\displaystyle PH=RE+PD$ $\displaystyle QH=QE+RD$ After squaring the two above equations, we get: $\displaystyle PH^{2}=RE^{2}+2\cdot RE\cdot PD + PD^{2}$ and $\displaystyle QH^{2}=QE^{2}+2\cdot QE\cdot RD + RD^{2}$. After the addition of two previous equations, we get: $\displaystyle PH^{2}+QH^{2}=RE^{2}+QE^{2}+PD^{2}+RD^{2}+2\cdot(R E\cdot PD + QE\cdot RD)$ Since $\displaystyle RE^{2}+QE^{2}=RQ^{2}$ and $\displaystyle PD^{2}+RD^{2}=PR^{2}$, we have: $\displaystyle PH^{2}+QH^{2}=RQ^{2}+PR^{2}+2\cdot(RE\cdot PD + QE\cdot RD)$, i.e: $\displaystyle PH^{2}+QH^{2}=(RQ+PR)^{2}2\cdot PR\cdot RQ+2\cdot(RE\cdot PD + QE\cdot RD)$. If points $\displaystyle P$, $\displaystyle R$ and $\displaystyle Q$ are collinear, we would have: $\displaystyle PH^{2}+QH^{2}=PQ^{2}$. So, that $\displaystyle P$, $\displaystyle Q$ and $\displaystyle R$ would be collinear, we have to prove that $\displaystyle PR\cdot RQ= RE\cdot PD + QE\cdot RD$. Here I stuck. 
September 14th, 2018, 01:00 PM  #3 
Member Joined: Jan 2018 From: Belgrade Posts: 50 Thanks: 2 
When I look better, last equation seems to be the same as $\displaystyle cos^{2}\alpha + sin^{2}\alpha =1$ (when the equation is devided by $\displaystyle PR\cdot RQ$), where $\displaystyle \alpha =\angle PRD$ which is equal to $\displaystyle \angle RQH$, when $\displaystyle P$, $\displaystyle R$ and $\displaystyle Q$ are collinear. In this moment, I'm feeling like I'm running my own tail. 

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