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