Two triangles and three collinear points

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Phlembac Adib Hasan
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Two triangles and three collinear points

Unread post by Phlembac Adib Hasan » Mon Nov 07, 2016 10:19 pm

We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear.
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joydip
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Re: Two triangles and three collinear points

Unread post by joydip » Mon Nov 21, 2016 11:44 pm

$AD$,$BE$ and $CF$ concur on the orthocenter $H$ of $\triangle DEF$. Let $D_0,E_0,F_0 $ be the projection of $D,E,F$ on $EF,FD $ and $DE$ respectively .Let $\alpha$ be the circumcircle of $\triangle DEF$.

Lemma : The perpendiculars to $BC, CA, AB$ through $ R,S,T $ respectively intersect at $H$

Proof : Let $EF$ meet $BC$ at $L$ .Then $(LB,LF;LH,LA)=-1 \Rightarrow ( D,D_0;H,A) =-1$. Let $F_1$ be the antipode of $F$.$F_1H$ meet $\alpha$ again at $F_2$ . $FF_2$ meet $DD_0$ at $D_3$ . Now , $F_1EHD$ is a parallelogram .Let $K$ be the midpoint of $DE$. Then $F_1,K,H $ are collinear . $\angle HF_2F=90 ^\circ $ .So $H,D_0,F,F_2,E_0$ lie on a circle ,say $ \psi $ .$\angle KE_0H= \angle KEH =\angle E_0FH$. So $KE_0$ is tangent to $ \psi $ .Similarly $KD_0$ is tangent to $\psi$. So $ F_2D_0H E_0$ is harmonic $ \Rightarrow (F E_ 0,FD_0 ; FH, FF_2 ) =(D,D_0;H,D_3) =-1$. So $A=D_3 \Rightarrow F_2=T$.So $HT \perp AT$ ,which proves the lemma.

As previous ,Let $E_1$ be the antipode of E. $FS$ intersect $ET$ at $Z$,$BS$ intersect $CT$ at $Y_1$. Applying pascal theorem on hexagon $EE_1SFF_1T$ we get $H,O,Z$ are collinear. Applying pappus theorem on hexagon $EBSFCT$ we get $H,Z,Y_1$ are collinear .So, $AR,BS,CT$ concur on $OH$.

Let the perpendiculars to $AC,AB$ through $E,F$ meet at $X_1$.$HT \perp AB$ and $HS \perp AC$. So the perpendicular bisectors of $TF$ and $SE$ goes through the midpoint of $HX_ 1$. Then O is midpoint of $HX_ 1$, which proves their concurrency on $OH$ ,completing the proof.
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