My Math Forum Prove that P, Q and R are collinear

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August 17th, 2018, 04:08 AM   #1
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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.
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September 14th, 2018, 12:34 PM   #2
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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.
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 Drz2016_VIII_z5_2.jpg (7.8 KB, 1 views)

 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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