## A Lemma?

- Thamim Zahin
**Posts:**98**Joined:**Wed Aug 03, 2016 5:42 pm

### A Lemma?

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, $OA \perp \lambda$.

1. Prove that, $OA \perp H_BH_C$.

2. Let a line $\lambda$ be tangent to $(HBC)$ at $H$. Prove that, $OA \perp \lambda$.

Last edited by Thamim Zahin on Sat Apr 15, 2017 1:24 pm, edited 3 times in total.

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### Re: A Lemma?

Define $H_BH_C$.

The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

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- Raiyan Jamil
**Posts:**138**Joined:**Fri Mar 29, 2013 3:49 pm

### Re: A Lemma?

$H_B$ is the feet of perpendicular from $B$ to $AC$ nd figure $H_C$ urself.dshasan wrote:Define $H_BH_C$.

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- Thanic Nur Samin
**Posts:**176**Joined:**Sun Dec 01, 2013 11:02 am

### Re: A Lemma?

This note might be relevant.

Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

Because destroying everything mindlessly isn't cool enough.

- Thanic Nur Samin
**Posts:**176**Joined:**Sun Dec 01, 2013 11:02 am

### Re: A Lemma?

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 $AO$.

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 $AO$.

Last edited by Thanic Nur Samin on Mon Apr 17, 2017 8:32 pm, edited 1 time in total.

Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

Because destroying everything mindlessly isn't cool enough.

- Mallika Prova
**Posts:**6**Joined:**Thu Dec 05, 2013 7:44 pm**Location:**Mymensingh,Bangladesh

### Re: A Lemma?

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 $EF\|H_bH_c$

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 $EF\|H_bH_c$

Last edited by Mallika Prova on Mon Apr 17, 2017 8:28 pm, edited 1 time in total.

**If you do what interests you , atleast one person is pleased.**- Mallika Prova
**Posts:**6**Joined:**Thu Dec 05, 2013 7:44 pm**Location:**Mymensingh,Bangladesh

### Re: A Lemma?

surely it wants to be $MN||AO$ ??Thanic Nur Samin wrote: $MN||AH$

**If you do what interests you , atleast one person is pleased.**- ahmedittihad
**Posts:**181**Joined:**Mon Mar 28, 2016 6:21 pm

### Re: A Lemma?

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

Frankly, my dear, I don't give a damn.

- Thanic Nur Samin
**Posts:**176**Joined:**Sun Dec 01, 2013 11:02 am

### Re: A Lemma?

Thanks I edited that.Mallika Prova wrote: surely it wants to be $MN||AO$ ??

Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

Because destroying everything mindlessly isn't cool enough.