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

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 .

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

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$ .

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$

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

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$.

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}$.

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