255. A line which is parallelly with OH.

by Virgil Nicula, Mar 20, 2011, 2:06 PM

Proposed problem. Let $ABC$ be an acute triangle with the circumcircle $w=C(O,R)$ and the orthocenter $H$ . Denote the reflections

$P$ , $Q$ , $R$ of the vertices $A$ , $B$ , $C$ in the sidelines $BC$ , $CA$ , $AB$ respectively. Denote $D\in CQ\cap BR$ . Prove that $PD\parallel OH$ .

Proof. Since the points $ R$ , $ Q$ are the reflections of the points $ C$ , $ B$ respectively w.r.t the lines $ AB$ , $ AC$ respectively results that the rays $ [BA$ , $ [CA$ are the bisectors of the

angles $ \widehat {CBR}$ , $ \widehat {BCQ}$ respectively , i.e. the point $ A$ is the $ D$ - exincenter of the triangle $ BDC$. Denote the diameter $ [AA_1$ of the circumcircle for the triangle $ ABC$. Since

$ A_1B\perp AB$ , $ A_1C\perp AC$ results that the point $ A_1$ is the incenter of the triangle $ BDC$ , i.e. the ray $ [DA_1$ is the bisector of the angle $ \widehat {BDC}$ . Therefore, $ A_1\in AD$ .

Since $ O\in AA_1$ results that the points $ A$ , $ O$ , $ A_1$ , $ D$ are collinearly. Since $ OB = OC$ results the point $ O$ is the intersection between the bisector of the segment $ [BC]$

and the bisector of the angle $ \widehat {BDC}$ . In conclusion, the quadrilateral $ OBDC$ is cyclically. Observe that $ m(\widehat {BDC}) = 180 - 2A$ $ \Longleftrightarrow$ $ m(\widehat {ADC}) = 90 - A$ $ \Longleftrightarrow$

$h_a = AD\sin (90 - A)$ $ \Longleftrightarrow$ $ 2Rh_a = AD\cdot 2R\cos A$ $ \Longleftrightarrow$ $ \frac {2R\cos A}{2h_a} = \frac {R}{AD}$ $ \Longleftrightarrow$ $ \frac {AH}{AP} = \frac {AO}{AD}$ $ \Longleftrightarrow$ $ HO\parallel PD$ .

An equivalent enunciation. Let $ ABC$ be a triangle. Denote the reflection $ P$ of the $ A$ - exincenter

$ I_a$ w.r.t. the sideline $ BC$. Prove that the line $ AP$ is parallelly with the Euler's line of the triangle $ BI_aC$ .

So maybe the our problem becomes more nicely and the its equivalent enunciation is more shortly !
This post has been edited 10 times. Last edited by Virgil Nicula, Nov 22, 2015, 9:53 AM

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