ΔABC is a triangle such that AB≠AC. The incircle of ΔABC touches BC,CA,AB at D,E,F respectively. H is a point on the segment EF such that DH⊥EF. Suppose AH⊥BC, prove that H is the orthocentre of ΔABC. Remark: the original question has missed the condition AB≠AC
Problem
Source: 11th CHKMO 2009
Tags: geometry, trigonometry, ratio, incenter, inradius, circumcircle, parallelogram
15.12.2008 06:50
Remark: can anybody provide a nicer proof for ∠HCD=90−∠C?
15.12.2008 08:01
brianchung11 wrote: ΔABC is a triangle such that AB≠AC. The incircle of ΔABC touches BC,CA,AB at D,E,F respectively. H is a point on the segment EF such that DH⊥EF. Suppose AH⊥BC, prove that H is the orthocentre of ΔABC. Remark: the original question has missed the condition AB≠AC An other solution . Let EF cuts BC at point T then (BCDT)=−1 .Because RH⊥EF so it is the angle bisector of ∠AHB . Therefore ∠BHP=∠AHQ . From this statement , it is easy to check that triangle AHQ is similarity to triangle BHP . So BH⊥AC , and H is the orthocentre of triangle ABC .
15.12.2008 14:49
http://www.mathlinks.ro/viewtopic.php?p=1321704&search_id=509318415#1321704 An almost equivalent problem here
18.12.2008 12:57
brianchung11 wrote: Let ABC be an acute triangle with the orthocenter H . The its incircle C(I,r) touches its sides in D∈(BC) , E∈(CA) , F∈(AB) respectively. Suppose AB≠AC and denote R∈AH∩EF . Prove that if DR⊥EF , then R≡H . Prove easily that (or is well-known) : m(∠EDF)=B+C2=90∘−A2 a.s.o. ; AH=2RcosA ; r=(p−a)tanA2=(p−b)tanB2=(p−c)tanC2 ; cosA+cosB+cosC=1+rR (1) . RFRE=AFAE⋅sin^RAFsin^RAE=sin(90∘−B)sin(90∘−C)=cosBcosC ⟹ RFRE=cosBcosC (2) . ARsin^AFR=AFsin^ARF⟹ARcosA2=p−asin(B+A2)⟹ AR=(p−a)cosA2sin(B+A2)= rtanA2⋅cosA2sin(B+A2)= r⋅cos2A2sin(B+A2)sinA2 ⟹ AR=r⋅1+cosAcosB+cosC (3) . ============================================= ▸ \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 .
02.01.2009 15:19
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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11.01.2009 15:26
why there is only one point T lie on AH there are two points symmetrical
12.01.2009 11:02
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.
13.01.2009 14:26
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}.
28.01.2011 05:13
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