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A Lemma?
Posted: Fri Apr 14, 2017 6:09 pm
by Thamim Zahin
In a $\triangle ABC$, let $H_A,H_B,H_C$ be the projection of $A,B,C$ on $BC,CA,AB$ respectively. And $O$ is the circumcenter
1. Prove that, $OA \perp H_BH_C$.
2. Let a line $\lambda$ be tangent to $(HBC)$ at $H$. Prove that, $OA \perp \lambda$.
Re: A Lemma?
Posted: Fri Apr 14, 2017 10:41 pm
by dshasan
Define $H_BH_C$.
Re: A Lemma?
Posted: Fri Apr 14, 2017 10:51 pm
by Raiyan Jamil
dshasan wrote:Define $H_BH_C$.
$H_B$ is the feet of perpendicular from $B$ to $AC$ nd figure $H_C$ urself.
Re: A Lemma?
Posted: Mon Apr 17, 2017 12:27 pm
by Thanic Nur Samin
This note might be relevant.
Re: A Lemma?
Posted: Mon Apr 17, 2017 12:40 pm
by Thanic Nur Samin
Also, an elegant proof for part 2 is presented below.
Take the homothety with center $A$ and ratio $\dfrac{1}{2}$. This sends $(HBC)$ to the nine point circle. Let the center of the nine point circle be $N$, and let the midpoint of $AH$ be $M$. Now, since $N$ is the midpoint of $OH$, $MN||AO$[typo edited], and since $H$ is sent to $M$, the tangent is perpendicular to $MN$, which is consequently parallel to $AO$. Thus the original tangent was also perpendicular to $AO$.
Re: A Lemma?
Posted: Mon Apr 17, 2017 7:52 pm
by Mallika Prova
a proof not elegant
1.
$\angle ABH_a =\angle AH_bH_c $ and
$\angle BAH_a =\angle H_bAO $ as $O$ and $H$ are isogonal conjugates
2.
definitely not elegant
let the tangent meet $AB$ and $BC$ at $E$,$F$
$\angle H_bHF=\angle EHB =\angle HCB=\angle HH_bH_c $
and gives $EF\|H_bH_c$
Re: A Lemma?
Posted: Mon Apr 17, 2017 8:01 pm
by Mallika Prova
Thanic Nur Samin wrote: $MN||AH$
surely it wants to be $MN||AO$ ??
Re: A Lemma?
Posted: Mon Apr 17, 2017 8:03 pm
by ahmedittihad
I think you have the wrong idea of elegance. To see what not elegant is, Thanic Nur Samin vaiya is your best choice.
Re: A Lemma?
Posted: Mon Apr 17, 2017 8:33 pm
by Thanic Nur Samin
Mallika Prova wrote:
surely it wants to be $MN||AO$ ??
Thanks
I edited that.