Prove H is the orthocentre from given condition

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Prove H is the orthocentre from given condition

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Source: 11th CHKMO 2009

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#1 h Dec 15, 2008, 4:19 AM • 1 Y

Y by Adventure10

$\Delta ABC$ is a triangle such that $AB \neq AC$ . The incircle of $\Delta ABC$ touches $BC, CA, AB$ at $D, E, F$ respectively. $H$ is a point on the segment $EF$ such that $DH \bot EF$ . Suppose $AH \bot BC$ , prove that $H$ is the orthocentre of $\Delta ABC$ .

Remark: the original question has missed the condition $AB \neq AC$

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#2 h Dec 15, 2008, 4:50 AM • 2 Y

Y by Adventure10, Mango247

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Remark: can anybody provide a nicer proof for $\angle HCD = 90 - \angle C$ ?

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TTsphn

#3 h Dec 15, 2008, 6:01 AM • 1 Y

Y by Adventure10

brianchung11 wrote:

$\Delta ABC$ is a triangle such that $AB \neq AC$ . The incircle of $\Delta ABC$ touches $BC, CA, AB$ at $D, E, F$ respectively. $H$ is a point on the segment $EF$ such that $DH \bot EF$ . Suppose $AH \bot BC$ , prove that $H$ is the orthocentre of $\Delta ABC$ .

Remark: the original question has missed the condition $AB \neq AC$

An other solution . Let $EF$ cuts $BC$ at point T then $(BCDT) = - 1$ .Because $RH\perp EF$ so it is the angle bisector of $\angle{AHB}$ . Therefore $\angle{BHP} = \angle{AHQ}$ . From this statement , it is easy to check that triangle AHQ is similarity to triangle BHP . So $BH\perp AC$ , and H is the orthocentre of triangle ABC .

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#4 h Dec 15, 2008, 12:49 PM • 2 Y

Y by Adventure10, Mango247

http://www.mathlinks.ro/viewtopic.php?p=1321704&search_id=509318415#1321704
An almost equivalent problem here

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#5 h Dec 18, 2008, 10:57 AM • 2 Y

Y by Adventure10, Mango247

brianchung11 wrote:

Let $ABC$ be an acute triangle with the orthocenter $H$ . The its incircle $C(I,r)$ touches

its sides in $D\in (BC)$ , $E\in (CA)$ , $F\in (AB)$ respectively. Suppose $AB\ne AC$

and denote $R\in AH\cap EF$ . Prove that if $DR\perp EF$ , then $R\equiv H$ .

Prove easily that (or is well-known) : $m(\angle EDF) = \frac {B + C}{2} = 90^{\circ} - \frac A2$ a.s.o. ; $AH = 2R\cos A$ ;

$r = (p - a)\tan\frac A2 = (p - b)\tan\frac B2 = (p - c)\tan\frac C2$ ; $\boxed {\cos A + \cos B + \cos C = 1 + \frac rR}\ \ (1)$ .

$\frac {RF}{RE} = \frac {AF}{AE}\cdot\frac {\sin\widehat {RAF}}{\sin\widehat {RAE}} = \frac {\sin\left(90^{\circ} - B\right)}{\sin\left(90^{\circ} - C\right)} = \frac {\cos B}{\cos C}$ $\implies$ $\boxed {\ \frac {RF}{RE} = \frac {\cos B}{\cos C}\ }\ \ (2)$ .

$\frac {AR}{\sin \widehat {AFR}} = \frac {AF}{\sin\widehat {ARF}}\implies\frac {AR}{\cos\frac A2} = \frac {p - a}{\sin\left(B + \frac A2\right)}\implies$ $AR = \frac {(p - a)\cos\frac A2}{\sin\left(B + \frac A2\right)} =$

$\frac {r}{\tan\frac A2}\cdot \frac {\cos\frac A2}{\sin\left(B + \frac A2\right)} =$ $r\cdot\frac {\cos^2\frac A2}{\sin\left(B + \frac A2\right)\sin\frac A2}$ $\implies$ $\boxed {\ AR = r\cdot\frac {1 + \cos A}{\cos B + \cos C}\ }\ \ (3)$ .

$= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =$

$\blacktriangleright\ DR\perp EF\implies$ $\frac {RF}{RE} = \frac {DF}{DE}\cdot\frac {\sin\widehat {RDF}}{\sin\widehat {RDE}}\stackrel {(2)}{\ \implies\ }\frac {\cos B}{\cos C} = \frac {2(p - b)\sin\frac B2}{2(p - c)\sin\frac C2}\cdot \frac {\sin\frac C2}{\sin \frac B2}$ $\implies$

$\frac {\cos B}{\cos C} = \frac {p - b}{p - c} = \frac {\tan\frac C2}{\tan\frac B2}$ $\implies$ $\cos B\tan\frac B2 = \cos C\tan\frac C2\stackrel {(B\ne C)}{\ \implies\ }\underline {1 + \cos A = \cos B + \cos C}$ .

Thus, $\left\|\begin{array}{c} (3)\implies\underline {AR = r} \\ \\ (1)\implies 2R\cos A = r\implies \underline {AH = r}\end{array}\right\|\ \implies\ AR = AH = r\ \implies\ R\equiv H$ .

Hence $\boxed {\ DR\perp EF\ \implies\ R\equiv H\ }$ . Observe that if $M$ is the midpoint of $[BC]$ , then $H\in MI$ .

Remark. If $B\ne C$ , then $1+\cos A=\cos B+\cos C\Longleftrightarrow$ $1+\sin B\sin C=\frac {\sin (B-C)}{\sin B-\sin C}\Longleftrightarrow$

$\sin\frac A2\cos\frac B2\cos\frac C2=\frac 12\Longleftrightarrow$ $(p-b)(p-c)=2Rr\Longleftrightarrow$ $OI=OD$ , where $O$ is the circumcenter of $\triangle ABC$ .

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#6 h Jan 2, 2009, 1:19 PM • 2 Y

Y by Adventure10, Mango247

We have:
* $\angle{HED}=\angle{DIB}=90-\frac{\angle{B}}{2}$
$\rightarrow \triangle{DHE} \sim \triangle{BDI} \rightarrow ID.HD=BD.HE$ (1)
* $\angle{HFD}=\angle{DIC}=90-\frac{\angle{B}}{2}$
$\rightarrow \triangle{DHF} \sim \triangle{CDI} \rightarrow ID.HD=CD.HF$ (2)
From (1) ,(2) we get : $\triangle{HEC} \sim \triangle{HFB} \rightarrow \angle{ABH}=\angle{ACH}$ (3)
$CH$ intersect $AB$ ; $BH$ intersect $AC$ at $P,Q$ ,respectively
$\rightarrow APQC$ is inscribed a circle.
We have:
* $\angle{HAB}=90-\angle{ABC}=90-\angle{AQP}$
$\rightarrow AH$ pass through circumcenter of $\triangle APQ$
* $\angle{AHP}=180-\angle{HAB}-\angle{APH}=90-\angle{PCB}=90-\angle{PQH}$
$\rightarrow AH$ pass through circumcenter of $\triangle HPQ$
But, there is only one point $T$ lie on $AH$ such that $TQ=TP$ ( $AB\neq AC$ )
$\rightarrow APHQ$ is inscribed a circle $\rightarrow \angle{APH}=\angle{AQH}=90$
Therefore, $H$ is orthocenter of triangle $ABC$

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#7 h Jan 11, 2009, 1:26 PM • 1 Y

Y by Adventure10

why there is only one point T lie on AH
there are two points symmetrical

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#8 h Jan 12, 2009, 9:02 AM • 2 Y

Y by Adventure10, Mango247

Another approach: Let $T$ be the intersection of $EF$ with $BC$ . We have that $(T, B, D, C)=-1$ , and since $DH \perp TH$ , the line $HD$ is the $H$ -angle bisector of triangle $BHC$ . Now, $AI \perp EF$ and so $AI \| DH$ . Combining this with $AH \perp BC$ and $ID \perp BC$ , we get that $AIDR$ is a parallelogram (where $I$ is the incenter of $ABC$ ). Consider now the homothety which maps $A$ to $R$ and $I$ to $D$ . Since in a triangle the circumcenter and orthocenter are isogonal conjugates, the $A$ , $R$ -circumcenter cevians of $ABC$ and $RBC$ are parallel and so the circumcenter $O$ of $ABC$ is mapped into the circumcenter $O'$ of $BRC$ . It follows that $AH = 2OM$ , where $M$ is the midpoint of $BC$ , and so $H$ is the orthocenter of $ABC$ .

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#9 h Jan 13, 2009, 12:26 PM • 2 Y

Y by Adventure10, Mango247

Once we have proven that $\angle{BHD}=\angle{CHD}$ , as $\angle{HFB}=\angle{HEC}$ , we have $\angle{ABH}=\angle{ACH}$ . Then reflect $C$ with respect to $AH$ , and call this new point $C'$ . Therefore $AHBC'$ is cyclic quadrilateral and $\angle{HBC}=\angle{HAC'}=\angle{HAC}=90-\angle{C}$ .

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#10 h Jan 28, 2011, 3:13 AM • 2 Y

Y by Adventure10, Mango247

Let's assume without loss of generality that AB < AC, Sea I incenter ABC then it is easy to see that AI is perpendicular to EF that AI / / DH also as ID is perpendicular to BD have ID / / AH so as DIAH is a parallelogram, then AH = ID, AI Pack = DH, < DBF = 2 < IBD = < EHL besides < EHD = 90 = < IDB then IBD and DEH are similar so EH/HD = ID / DB = AH/DB from here
EH/HD = AH / DB i.e. EH / AH = HD /DB and < EHA = < HDB, DHB and EHA are therefore similar < DHB = <AEH= < AFH = x then < EAF = 180 - 2 x now prove that < CHD = x continuing EOF until that Court to CB in M luiego is known that C, D, B and M are in Tetrad harmony and as < DHM = 90 get < DHB = < CHD = x this means that H meets that < CAB + < CHB = 180 and H is the height so as the orthocenter only conclude that H is the orthocenter. THIS COMPLETES THE SOLUTION

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