Where I visualize cyclic ness?

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Fm Jakaria
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Where I visualize cyclic ness?

Unread post by Fm Jakaria » Mon Sep 16, 2013 12:39 am

Let $ABC$ be an acute angled triangle with altitudes $AD, BE$ and $CF$. Suppose $EF$ intersects $BC$ at $P$. Let the line through $D$ parallel to $EF$ intersect $AB$ at $R$ and $AC$ at $Q$. Prove that the circumcircle of triangle $PQR$ passes through the midpoint of side $BC$.
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photon
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Re: Where I visualize cyclic ness?

Unread post by photon » Sun Sep 29, 2013 10:23 pm

$\angle RQE = \angle QEP = \angle AEF = \angle ABC$
$\angle RQC = \angle RBC$ , $B,R,C,Q$ are cycic . $\therefore RD.DQ=BD.DC$
Let M be the midpoint of $BC$ . $M,D,E,F$ lie in nine-point circle . so , $PE.PF=PD.PM$
$\displaystyle \Rightarrow PC.PB = PD.PM$ [$PE.PF=PC.PB$ as $B,F,E,C$ are cyclic]
$\displaystyle \Rightarrow \frac{PM}{PC}=\frac{PB}{PD}$ $\displaystyle \Rightarrow \frac{MC}{PC}=\frac{BD}{PD}$
$\displaystyle \Rightarrow \frac{BM}{PC}=\frac{BD}{PD}............(1)$
$\displaystyle \Rightarrow \frac{BM}{BD}=\frac{PC}{PD} \Rightarrow \frac{BD}{BM}=\frac{PD}{PC} \Rightarrow \frac{MD}{BM}=\frac{CD}{PC} \Rightarrow \frac{MD}{CD}=\frac{BM}{PC}............(2)$
from (1),(2) - $\displaystyle \frac{MD}{CD}=\frac{BD}{PD}$
$MD.PD=BD.CD \Rightarrow MD.PD=RD.DQ$
so $M,Q,P,R$ are cyclic.
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*Mahi*
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Re: Where I visualize cyclic ness?

Unread post by *Mahi* » Wed Oct 02, 2013 11:06 am

Previously posted: viewtopic.php?f=13&t=2622 here.
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zadid xcalibured
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Re: Where I visualize cyclic ness?

Unread post by zadid xcalibured » Mon Oct 14, 2013 6:52 pm

:mrgreen: :mrgreen: :mrgreen: :mrgreen:

tanmoy
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Re: Where I visualize cyclic ness?

Unread post by tanmoy » Sat Oct 19, 2013 11:37 am

why $\angle AEF=\angle ABC$?

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