BdMO Online Forum

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, $OA \perp \lambda$.

Define $H_BH_C$.

dshasan wrote:Define $H_BH_C$.

$H_B$ is the feet of perpendicular from $B$ to $AC$ nd figure $H_C$ urself.

This note might be relevant.

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 $AO$.

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 $EF\|H_bH_c$

Thanic Nur Samin wrote: $MN||AH$

surely it wants to be $MN||AO$ ??

I think you have the wrong idea of elegance. To see what not elegant is, Thanic Nur Samin vaiya is your best choice.

Mallika Prova wrote: surely it wants to be $MN||AO$ ??

Thanks I edited that.