Assume WLOG $AC \le AB$ Easy to see $SI=SC$ which implies $LI=LC$. And since $P$ is the reflection of $I$ through $KL$ we get $LI=LP$. Hence $L$ is the circumcenter of $\triangle IPC$ Also $K$ is the circumcenter of $\triangle IPB$. Call $\angle BAC= 4 \alpha$. $\angle ABC= 2 \phi$ $\angle ACB= 2 \beta$ Also $\angle PIS= x$ Just notice that $\angle SIL= \angle SCL= \alpha= \angle SBK= \angle SIK$. Hence $\angle KIL = 2 \alpha = \angle KPL$ Do the angle work is $\triangle IPC$ and since $LP= LC$ we get $\angle LPC= \phi - x$. Do the angle work in $\triangle IPB$ and since $KB= KP$ we get $\angle BPK= \beta +x$ Thus $\angle BPC= 2 \alpha + \phi + \beta +x-x=90$. Sorry if I didn't elaborate, but I didn't think it was necessary

My solution: Lemma: Let $ABC$ be a triangle such that $AB=AC$ let $D$ be a point in $BC$ and $M$ is the midpoint of $BC$ and let $I$ and $J$ the incenter of $\triangle ABD$ and $\triangle ACD$ $\Longrightarrow$ $I,D,M,J$ are concyclic Proof: It's well know by comparing the ceva's trigonometric in $\triangle ABD$ and $\triangle ACD$ we get $\angle IMB=\angle AMJ$ and $\angle IMA=\angle IMC$ $\Longrightarrow$ $\angle IMJ=90^{\circ}$, but $\angle IDM=90^{\circ}$ $\Longrightarrow$ $I,D,M,J$ are concyclic In the problem: Let $M$ the midpoint of $BC$ and let $Q=AI\cap KL$ and let $N$ the midpoint of $KL$. By lemma we get $KLDM$ are cyclic and $N$ is center of $\odot (KLDM)$ since $DK$ is bisector of $\angle BDS$ $\Longrightarrow$ $\angle KDS=\angle BDK=\angle MLK=\angle NML$ since $\angle DML+\angle DKL=180^{\circ}$ and $\angle KDS=\angle NML$ we get $\angle DQK=\angle DMN$ $\Longrightarrow$ $DMNQ$ is cyclic $\Longrightarrow$ $\angle MQN=\angle NQS$ $\Longrightarrow$ $KL$ is bisector of $\angle MQS$ $\Longrightarrow$ $P\in MQ$, but the bisectors of $\angle MQS$, $\angle QSL$ and $\angle SCD$ are concurrent $\Longrightarrow$ $QMCS$ is a quadrilateral circumscribed $\Longrightarrow$ $MQ+CS=QS+MC...(1)$ but $CS=BS=IS=QS+ID...(2)$ hence by $(1)$ and $(2)$ we get $MP=MC=MB$ $\Longrightarrow$ $\angle BPC=90^{\circ}$

The central claim here is that $\angle KIL+\angle KSL=90^\circ$. Let $M$ and $N$ be the $S$-excenters of $SDB$ and $SDC$ respectively. We know that $SK\cdot SM=SL\cdot SN=SD\cdot SB=SD\cdot SC=SD\cdot SI$. Note that $M$, $D$, and $L$ are collinear, so inversion about $S$ with radius $\sqrt{SD\cdot SI}$ results in the fact that $SKIN$ is cyclic. Similarly, $SLIM$ is cyclic, so $\angle KSL=\angle MSI+\angle NSI=\angle MLI+\angle NKI=90^\circ-\angle KIL$ as desired. Now, note that $B$, $P$, and $C$ are the reflections of $I$ across $SK$, $KL$, and $LS$ respectively. Hence, it is a right triangle iff the pedal triangle of $I$ with respect to $KSL$ is. But the pedal triangle of $I$ is similar to the circumcevian triangle of $I$, so the problem is equivalent to showing that $IK$ and $IL$ intersect the circumcircle of $KSL$ again at the endpoints of a diameter. But this is clear from $\angle KIL+\angle KSL=90^\circ$ so we are done.

Good problem! 1. $BKDI$ is cyclic. Proof. Trivial. $\angle BKD = 90^\circ + \angle BSD/2 = 180^\circ - \angle BID$ by Fact 5, $BS = IS$. 2. $BK = IK$. Proof. Trivial. $K$ lies on angle bisector of $\angle BSI$, and we mentioned $BS = IS$. 3. $BP$ passes through intersection of circle $BKDI$ and $KL$. Proof. Trivial. Read (2) first! The rest is angle chasing. 4. $\angle CBP = \angle DKL$. Proof. Trivial. Just use cyclic quad! 5. $\angle BCP = \angle DLK$. Proof. Similar to steps 1, 2, 3, 4. Trivial, eh? 6. $\angle BPC = 90^\circ$. Proof. Trivial. $KDL$ and $BPC$ are similar and $\angle KDL = 90^\circ$ is obvious from internal/external bisector theorem. 6 Trivial statements = 1 Complete Proof

Wait excuse me I am bad at math I can do this problem in $3$ trivial steps! 1. $BK = KI = KJ$. Proof. $BK = IK$ is (2) of above proof, and $KI = KJ$ is trivial property of reflections. 2. $\angle BPC = 1/2 \angle BKI + 1/2 \angle CLI$ Proof. $K$ is circumcenter of $BIP$ and similarly $L$ is circumcenter of $CIP$. 3. $\angle BKI + \angle CLI = 180^\circ$. Proof. $\angle BKI = 2 (90^\circ - \angle BDS/2)$ and $\angle CLI = 2 (90^\circ - \angle CDS/2)$ because $SK$ bisects $\angle BSD$ and $\angle BKS = 90^\circ + \angle BDS/2$

Since $\widehat{KDS}=\widehat{\frac{A}{4}}+\widehat{\frac{B}{2} } $ and $ \widehat{LDS}=\widehat{\frac{A}{4}}+\widehat{\frac{C}{2}} $ then $KDIB,LDIC$ are cyclic so $KI=KB,LI=LC$ because they subtend supplementary angles thus $\widehat{BPC}=\widehat{BPK}+\widehat{KPL}+\widehat{LPC}=\widehat{BKP}+\widehat{KIL}+\widehat{LCP}$ but $\widehat{KIL}=\widehat{KID}+\widehat{DIL}=\widehat{KBD}+\widehat{DCL}$ therefore $\widehat{BPC}=\widehat{PBC}+\widehat{PCB}=\frac{\pi}{2}$. R HAS

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$BKDI$, $CLDI$ cyclic + $K$, $L$ center of $BIP$, $CIP$ -> $\angle BPC = \frac{1}{2} (\angle BKI + \angle CLI) = \frac{1}{2} (\angle BDI + \angle CDI) = 90$

The author of this beautiful problem is Burii. Draw all four tangent lines to both incircles and denote the tangency and intersection points like in the picture. [asy][asy]/* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(9.62cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.59, xmax = 5.03, ymin = -2.18, ymax = 5.86; /* image dimensions */ /* draw figures */ draw(circle((0.6,2.03), 3.57)); draw((xmin, -2.39*xmin-0.21)--(xmax, -2.39*xmax-0.21)); /* line */ draw((xmin, -0.01*xmin + 1.11)--(xmax, -0.01*xmax + 1.11)); /* line */ draw(circle((1,0.07), 1.03)); draw((xmin, 0.96*xmin + 0.53)--(xmax, 0.96*xmax + 0.53)); /* line */ draw((-1.98,4.51)--(-2.86,1.14)); draw((4.04,1.06)--(-1.98,4.51)); draw((4.04,1.06)--(0.56,-1.54)); draw((0.56,-1.54)--(-2.86,1.14)); draw(circle((-0.98,0.48), 0.64)); draw((xmin, -0.21*xmin + 0.28)--(xmax, -0.21*xmax + 0.28)); /* line */ draw((xmin, -0.42*xmin-0.63)--(xmax, -0.42*xmax-0.63)); /* line */ /* dots and labels */ dot((-1.98,4.51),dotstyle); label("$A$", (-2,4.75), NE * labelscalefactor); dot((-2.86,1.14),dotstyle); label("$B$", (-2.72,1.26), NE * labelscalefactor); dot((4.04,1.06),dotstyle); label("$C$", (4.12,0.74), NE * labelscalefactor); dot((0.56,-1.54),dotstyle); label("$S$", (0.3,-1.88), NE * labelscalefactor); dot((-1.12,2.47),dotstyle); label("$I$", (-1.05,2.57), NE * labelscalefactor); dot((-0.98,0.48),dotstyle); label("$K$", (-0.92,0.58), NE * labelscalefactor); dot((1,0.07),dotstyle); label("$L$", (1.07,0.17), NE * labelscalefactor); dot((1.01,1.1),dotstyle); label("$U$", (1.08,1.2), NE * labelscalefactor); dot((-1.9,-1.29),dotstyle); label("$P$", (-2.13,-1.25), NE * labelscalefactor); dot((0.59,1.1),dotstyle); label("$M$", (0.43,1.26), NE * labelscalefactor); dot((-0.55,1.11),dotstyle); label("$D$", (-0.48,1.22), NE * labelscalefactor); dot((0.29,0.81),dotstyle); label("$V$", (0.38,0.56), NE * labelscalefactor); dot((-0.97,1.12),dotstyle); label("$E$", (-1.04,1.23), NE * labelscalefactor); dot((-0.39,0.73),dotstyle); label("$H$", (-0.32,0.82), NE * labelscalefactor); dot((-0.54,0.02),dotstyle); label("$J$", (-0.4,-0.14), NE * labelscalefactor); dot((-1.38,-0.02),dotstyle); label("$N$", (-1.31,0.07), NE * labelscalefactor); dot((1.61,-0.75),dotstyle); label("$O$", (1.72,-0.92), NE * labelscalefactor); dot((0.05,-0.33),dotstyle); label("$Q$", (0.12,-0.24), NE * labelscalefactor); dot((0.21,-0.72),dotstyle); label("$R$", (0.28,-0.61), NE * labelscalefactor); dot((-4.25,1.16),dotstyle); label("$T$", (-4.18,1.25), NE * labelscalefactor); dot((0.6,-0.88),dotstyle); label("$W$", (0.66,-0.77), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] First we can prove like in the solutions above that $B,C$ are reflections of $I$ with respect to $SK, SL$. Using symmetries we get $PJ=IH=BN=BE$ and $PV=IQ=CO=CU$. Observe that $$MU=RW=\frac{DT+RD-RT}{2}=DE.$$It follows that $$MC=MU+UC=DE+UC=\frac{DB+DS-SB}{2} + \frac{CD+CS-DS}{2} = \frac{BC}{2},$$therefore $M$ is the midpoint of $BC$. Now, $$2PM=PJ+JM+PV+VM=BE+ME+CU+MU=BC.$$This implies that $\angle BPC=90^\circ$.

Dear Mathlinkers, have a look at http://www.artofproblemsolving.com/community/c6t48f6h1311529_geometry Sincerely Jean-Louis

Here is my solution for this problem Solution We have: $K$ is incenter of $\triangle$ $BSD$ So: $SK$ passes through midpoint of $\stackrel\frown{AB}$ which not contain $C$ Hence: $PK$ is perpendicular bisector of $BI$ or $KB$ = $KI$ We also have: $P$ is reflection of $I$ through $KL$ then: $KI$ = $KP$ It leads to: $K$ is circumcenter of $\triangle$ $BPI$ So: $\widehat{BPI}$ = $\dfrac{1}{2}$ $\widehat{BKI}$ = $\widehat{KBS}$ + $\widehat{KSB}$ = $\dfrac{\widehat{BAC}}{4}$ + $\dfrac{\widehat{ACB}}{2}$ Similarly: $\widehat{CPI}$ = $\dfrac{\widehat{BAC}}{4}$ + $\dfrac{\widehat{ABC}}{2}$ Hence: $\widehat{BPC}$ = $\widehat{BPI}$ + $\widehat{CPI}$ = $\dfrac{\widehat{BAC}}{4}$ + $\dfrac{\widehat{ACB}}{2}$ + $\dfrac{\widehat{BAC}}{4}$ + $\dfrac{\widehat{ABC}}{2}$ = $\dfrac{\widehat{BAC} + \widehat{ABC} + \widehat{ACB}}{2}$ = $90^0$ Then: $PB$ $\perp$ $PC$

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By a lemma on incenter, $SB=SI$ , and since $SK$ bisects $\angle BSD $, we obtain that $$\angle KCD=\angle KBS=\angle KID$$ implying $KBID$ is cyclic. Similarly, so is $DICL$. Note that $KB=KI=KP$ and $LC=LI=LP$. Hence, $$\angle KPL=\angle KIL=\angle KID+\angle DIL=\angle KBC+\angle LBC=\angle KPB+\angle LPC.$$ After some chasing on angles in $\triangle BPC$, we are done.

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Notice $$180-\angle KDI=\angle KDS=\tfrac{1}{2}\angle BDS=\tfrac{1}{2}ABS=\angle IBK$$so $IBKD$ and similarly $ICLD$ are cyclic. Since $SB=IS,$ we know $KB=KI.$ We see $KI=KP$ by reflection, so $K$ is the circumcenter of $BIP.$ Similarly, $L$ is the circumcenter of $CIP.$ Hence, $$\angle BPC=\angle BPI+\angle CPI=\tfrac{1}{2}(\angle BKI+\angle CLI)=\tfrac{1}{2}(\angle IDB+\angle IDC)=90.$$$\square$

P3? This suits P1/ easy P2. Just notice $\measuredangle BKD = 90 ^\circ+ \frac{C}{2} = 180^\circ - \measuredangle BID$, hence $BKID$ is cyclic, similarly $CLID$ is cyclic. $$\measuredangle BPC = \measuredangle BPI + \measuredangle CPI = \frac{\measuredangle BKI + \measuredangle CLI}{2}=\frac{\measuredangle BDI + \measuredangle CDI}{2}=90^\circ$$

$\textbf{Lemma:}$ $ABC$ is a triangle with orthocenter $H$. $BH$ and $CH$ intersect $(ABC)$ at $D,E$ respectively and $DE$ intersects $AH$ at $F$. Let $K,L$ be the incenters of $AFE$ and $AFD$. If $P$ is the reflection of $H$ according to $KL$, then $\angle DPE=90$. \[\angle FKE=90+\frac{\angle FAE}{2}=180-\angle B=180-\angle EHF\implies H,F,K,E \ \text{are cyclic}\]\[\angle DLH=90+\frac{\angle DAH}{2}=180-\angle C=180-\angle FHD \implies H,F,L,D \ \text{are cyclic}\]Let $KL\cap (EHF)=R, KL\cap (DHF)=S,ER\cap DS=Q$. \[\angle QRK=\angle EFK=\angle KFA=\angle KRH \ \text{and} \ \angle LSQ=\angle LFD=\angle AFL=\angle HSL\]Hence, $Q$ is the reflection of $H$ according to $KL$. Also \[\angle SQR=\angle RHS=\angle RHF+\angle FHS=\angle KFL+\angle FLK=90\]Which gives the desired conclusion.$\square$

By replacing

we get $\angle BPC=90$ as desired.$\blacksquare$

As $SB = SI$ and $SK$ bisects $\angle BSI$, we have $KB=KI=KP$, so \[\angle BPI = \frac 12 \angle BKI = 180^\circ - \angle BKS = 90^\circ - \frac 12 \angle BDS\]and similarly $\angle CPI = 90^\circ - \frac 12 \angle CDS$, so $\angle BPC = 180^\circ - 90^\circ = 90^\circ$.

bin_sherlo wrote: $\textbf{Lemma:}$ $ABC$ is a triangle with orthocenter $H$. $BH$ and $CH$ intersect $(ABC)$ at $D,E$ respectively and $DE$ intersects $AH$ at $F$. Let $K,L$ be the incenters of $AFE$ and $AFD$. If $P$ is the reflection of $H$ according to $KL$, then $\angle DPE=90$. \[\angle FKE=90+\frac{\angle FAE}{2}=180-\angle B=180-\angle EHF\implies H,F,K,E \ \text{are cyclic}\]\[\angle DLH=90+\frac{\angle DAH}{2}=180-\angle C=180-\angle FHD \implies H,F,L,D \ \text{are cyclic}\]Let $KL\cap (EHF)=R, KL\cap (DHF)=S,ER\cap DS=Q$. \[\angle QRK=\angle EFK=\angle KFA=\angle KRH \ \text{and} \ \angle LSQ=\angle LFD=\angle AFL=\angle HSL\]Hence, $Q$ is the reflection of $H$ according to $KL$. Also \[\angle SQR=\angle RHS=\angle RHF+\angle FHS=\angle KFL+\angle FLK=90\]Which gives the desired conclusion.$\square$

By replacing

we get $\angle BPC=90$ as desired.$\blacksquare$ How can you say $180-\angle B=180-\angle EHF$?

Maksat_B wrote: bin_sherlo wrote: $\textbf{Lemma:}$ $ABC$ is a triangle with orthocenter $H$. $BH$ and $CH$ intersect $(ABC)$ at $D,E$ respectively and $DE$ intersects $AH$ at $F$. Let $K,L$ be the incenters of $AFE$ and $AFD$. If $P$ is the reflection of $H$ according to $KL$, then $\angle DPE=90$. \[\angle FKE=90+\frac{\angle FAE}{2}=180-\angle B=180-\angle EHF\implies H,F,K,E \ \text{are cyclic}\]\[\angle DLH=90+\frac{\angle DAH}{2}=180-\angle C=180-\angle FHD \implies H,F,L,D \ \text{are cyclic}\]Let $KL\cap (EHF)=R, KL\cap (DHF)=S,ER\cap DS=Q$. \[\angle QRK=\angle EFK=\angle KFA=\angle KRH \ \text{and} \ \angle LSQ=\angle LFD=\angle AFL=\angle HSL\]Hence, $Q$ is the reflection of $H$ according to $KL$. Also \[\angle SQR=\angle RHS=\angle RHF+\angle FHS=\angle KFL+\angle FLK=90\]Which gives the desired conclusion.$\square$

By replacing

we get $\angle BPC=90$ as desired.$\blacksquare$ How can you say $180-\angle B=180-\angle EHF ?$

bin_sherlo wrote: $\textbf{Lemma:}$ $ABC$ is a triangle with orthocenter $H$. $BH$ and $CH$ intersect $(ABC)$ at $D,E$ respectively and $DE$ intersects $AH$ at $F$. Let $K,L$ be the incenters of $AFE$ and $AFD$. If $P$ is the reflection of $H$ according to $KL$, then $\angle DPE=90$. \[\angle FKE=90+\frac{\angle FAE}{2}=180-\angle B=180-\angle EHF\implies H,F,K,E \ \text{are cyclic}\]\[\angle DLH=90+\frac{\angle DAH}{2}=180-\angle C=180-\angle FHD \implies H,F,L,D \ \text{are cyclic}\]Let $KL\cap (EHF)=R, KL\cap (DHF)=S,ER\cap DS=Q$. \[\angle QRK=\angle EFK=\angle KFA=\angle KRH \ \text{and} \ \angle LSQ=\angle LFD=\angle AFL=\angle HSL\]Hence, $Q$ is the reflection of $H$ according to $KL$. Also \[\angle SQR=\angle RHS=\angle RHF+\angle FHS=\angle KFL+\angle FLK=90\]Which gives the desired conclusion.$\square$

By replacing

we get $\angle BPC=90$ as desired.$\blacksquare$ Also you can't replace points $H and I$

Maksat_B wrote: bin_sherlo wrote: $\textbf{Lemma:}$ $ABC$ is a triangle with orthocenter $H$. $BH$ and $CH$ intersect $(ABC)$ at $D,E$ respectively and $DE$ intersects $AH$ at $F$. Let $K,L$ be the incenters of $AFE$ and $AFD$. If $P$ is the reflection of $H$ according to $KL$, then $\angle DPE=90$. \[\angle FKE=90+\frac{\angle FAE}{2}=180-\angle B=180-\angle EHF\implies H,F,K,E \ \text{are cyclic}\]\[\angle DLH=90+\frac{\angle DAH}{2}=180-\angle C=180-\angle FHD \implies H,F,L,D \ \text{are cyclic}\]Let $KL\cap (EHF)=R, KL\cap (DHF)=S,ER\cap DS=Q$. \[\angle QRK=\angle EFK=\angle KFA=\angle KRH \ \text{and} \ \angle LSQ=\angle LFD=\angle AFL=\angle HSL\]Hence, $Q$ is the reflection of $H$ according to $KL$. Also \[\angle SQR=\angle RHS=\angle RHF+\angle FHS=\angle KFL+\angle FLK=90\]Which gives the desired conclusion.$\square$

By replacing

we get $\angle BPC=90$ as desired.$\blacksquare$ How can you say $180-\angle B=180-\angle EHF$? Because $C,H,E$ are collinear and $\angle EHF=\angle EHA=\angle HCA+\angle CAH=\angle B$

Maksat_B wrote: bin_sherlo wrote: $\textbf{Lemma:}$ $ABC$ is a triangle with orthocenter $H$. $BH$ and $CH$ intersect $(ABC)$ at $D,E$ respectively and $DE$ intersects $AH$ at $F$. Let $K,L$ be the incenters of $AFE$ and $AFD$. If $P$ is the reflection of $H$ according to $KL$, then $\angle DPE=90$. \[\angle FKE=90+\frac{\angle FAE}{2}=180-\angle B=180-\angle EHF\implies H,F,K,E \ \text{are cyclic}\]\[\angle DLH=90+\frac{\angle DAH}{2}=180-\angle C=180-\angle FHD \implies H,F,L,D \ \text{are cyclic}\]Let $KL\cap (EHF)=R, KL\cap (DHF)=S,ER\cap DS=Q$. \[\angle QRK=\angle EFK=\angle KFA=\angle KRH \ \text{and} \ \angle LSQ=\angle LFD=\angle AFL=\angle HSL\]Hence, $Q$ is the reflection of $H$ according to $KL$. Also \[\angle SQR=\angle RHS=\angle RHF+\angle FHS=\angle KFL+\angle FLK=90\]Which gives the desired conclusion.$\square$

By replacing

we get $\angle BPC=90$ as desired.$\blacksquare$ Also you can't replace points $H and I$ I can replace, because $I$ is the orthocenter of the triangle constructed by midpoints of arcs $AB,BC,CA$?

bin_sherlo wrote: Maksat_B wrote: bin_sherlo wrote: $\textbf{Lemma:}$ $ABC$ is a triangle with orthocenter $H$. $BH$ and $CH$ intersect $(ABC)$ at $D,E$ respectively and $DE$ intersects $AH$ at $F$. Let $K,L$ be the incenters of $AFE$ and $AFD$. If $P$ is the reflection of $H$ according to $KL$, then $\angle DPE=90$. \[\angle FKE=90+\frac{\angle FAE}{2}=180-\angle B=180-\angle EHF\implies H,F,K,E \ \text{are cyclic}\]\[\angle DLH=90+\frac{\angle DAH}{2}=180-\angle C=180-\angle FHD \implies H,F,L,D \ \text{are cyclic}\]Let $KL\cap (EHF)=R, KL\cap (DHF)=S,ER\cap DS=Q$. \[\angle QRK=\angle EFK=\angle KFA=\angle KRH \ \text{and} \ \angle LSQ=\angle LFD=\angle AFL=\angle HSL\]Hence, $Q$ is the reflection of $H$ according to $KL$. Also \[\angle SQR=\angle RHS=\angle RHF+\angle FHS=\angle KFL+\angle FLK=90\]Which gives the desired conclusion.$\square$

By replacing

we get $\angle BPC=90$ as desired.$\blacksquare$ How can you say $180-\angle B=180-\angle EHF$? Because $C,H,E$ are collinear and $\angle EHF=\angle EHA=\angle HCA+\angle CAH=\angle B$ Sorry, I've made an obvious mistake, it's right

bin_sherlo wrote: Maksat_B wrote: bin_sherlo wrote: $\textbf{Lemma:}$ $ABC$ is a triangle with orthocenter $H$. $BH$ and $CH$ intersect $(ABC)$ at $D,E$ respectively and $DE$ intersects $AH$ at $F$. Let $K,L$ be the incenters of $AFE$ and $AFD$. If $P$ is the reflection of $H$ according to $KL$, then $\angle DPE=90$. \[\angle FKE=90+\frac{\angle FAE}{2}=180-\angle B=180-\angle EHF\implies H,F,K,E \ \text{are cyclic}\]\[\angle DLH=90+\frac{\angle DAH}{2}=180-\angle C=180-\angle FHD \implies H,F,L,D \ \text{are cyclic}\]Let $KL\cap (EHF)=R, KL\cap (DHF)=S,ER\cap DS=Q$. \[\angle QRK=\angle EFK=\angle KFA=\angle KRH \ \text{and} \ \angle LSQ=\angle LFD=\angle AFL=\angle HSL\]Hence, $Q$ is the reflection of $H$ according to $KL$. Also \[\angle SQR=\angle RHS=\angle RHF+\angle FHS=\angle KFL+\angle FLK=90\]Which gives the desired conclusion.$\square$

By replacing

we get $\angle BPC=90$ as desired.$\blacksquare$ Also you can't replace points $H and I$ I can replace, because $I$ is the orthocenter of the triangle constructed by midpoints of rays $AB,BC,CA$? You mean $I$ is the orthocenter of the median triangle of $ABC$ ?

@above Not median triangle. If $X,Y,Z$ are midpoints of arcs $BC,CA,AB$ in $(ABC)$, then $I$ is the orthocenter of $XYZ$. When you apply lemma with triangle $XYZ$ with orthocenter $I$, you get the desired result.