Note that if $D, E$ are the feet of the altitudes from $A, B$, then $\angle{SAB} = \angle{SPE}$ and $\angle{PES} = \angle{ABS} \implies \triangle{SBA} \sim \triangle{SEP}$, and their sides are in the ratio $\frac{SE}{SB}$. If $W$ is the midpoint of $CH$ then by Radical Center on $(CH), (AC), (AB)$, $AS, CH, DE$ concur at $X$ which by Brocard is the orthocenter of $\triangle{AWB}$, and $B, S, W$ are collinear. This means $\angle{BWX} = \angle{BAS} = \angle{BES} \implies \frac{SE}{SB} = \frac{\sin \angle{SBE}}{\sin \angle{BES}} = \frac{WH}{BH}$, and since $BC' = \frac{EP}{\sin \angle{C}}$, $\frac{BC'}{AB} = \frac{WH}{BH \sin \angle{C}}$. Let point $U$ be on $CH$ such that $SC \parallel BU$; then it suffices to show that $\triangle{ABC'} \sim \triangle{UCB}$ since that would imply $\angle{C'AB} = \angle{BUC} \implies AC' \perp BU \implies SC \cap AC' \in (AC)$. Since $\angle{UCB} = \angle{ABC'}$, it suffices to show that $\frac{BC}{CU} = \frac{BC'}{AB}$. Since $\triangle{WSC} \sim \triangle{WBU}$, $CU = WC \cdot \frac{BS}{SW}$, and $BH \sin \angle{C} = BD$, so it suffices to show that $BC \cdot BD = WC^2 \cdot \frac{BS}{SW}$. Since $B = S^*$ inverting about $(CH)$, $WC^2 = WS \cdot WB \implies \frac{WC^2}{SW} = WB$, and so it suffices to show that $BC \cdot BD = BS \cdot BW$, which is clearly true since $\angle{BWX} = \angle{BAX} = \angle{SDC} \implies SWCD$ is cyclic. $\square$

Let $X=(AC)\cap CS$. We’ll prove that $BXPC$ is cyclic. Note that $\angle APS=\angle BAS=\angle SDC$ so $SDCP$ is cyclic. Now see that $\angle ASX=\pi-\angle ASC=\pi-\angle AHC=\angle B$. Hence the triangles $ABD$ and $ASX$ are similar due to right angle. Also, $\angle XDA=\angle XCA=\angle SCP=\angle SDP$ and $\angle AXD=\pi-\angle C=\angle PSD$. Thus the triangles $PSD$ and $AXD$ are similar. From the pais of similar triangles, $$\frac{BD}{XS}=\frac{AD}{AX}=\frac{DP}{SP}$$Since $DSPC$ is cyclic, we get $\angle BDP=\angle XSP$ so the triangles $BDP$ and $XSP$ are cyclic. Finally, from the last pair of similar triangles, we get $\angle PXC=\angle PXS=\angle PBD=\angle PBC$ so $BXPC$ is cyclic.

I've told almost every participant that all pervious years of Iran MO 3rd round had a problem which was solvable by the properties of Humpty-Dumpty points so probably this year would not be an exception. So I forced myself to solve this problem with those properties and thankfully it was possible. The main observation is that since in any triangle $\triangle ABC$ , If $\chi$ is $A$-humpty point , then $\chi$ lies on $(BHC)$ and $\angle H \chi A = 90$.So since $A$ is the orthocenter of $\triangle BHC$ , $S$ is the $B$-humpty point of $\triangle BHC$ and by another property of $\chi$ , $\frac{BH}{BC}=\frac{SH}{SC}$, the rest is bash: Let $K$ be the intersection of the perpendicular through $A$ to $AC$ and the circle with diameter $(AB)$. So $\angle BKS= \angle BAS= \angle SPA$ and since $BK || AC$ , $K,S,P$ are collinear. So since $\angle KAP=90$ , $\triangle{KAP} \sim \triangle{ASB}$ and we'll get $AP = AK.\frac{AS}{BS}$.Also , by the sine plotemy , it's enough to show that if $J=CS \cap (AC)$ and $\alpha = \angle SCA$ : $$CP =? \frac{b.cos \alpha . sin\angle C - a.sin\alpha }{sin \angle C - \alpha }$$So it's enough to show that: $$b - \frac{b.cos \alpha . sin\angle C - a.sin \alpha }{sin \angle C - \alpha } = AK.\frac{AS}{BS}$$$$b.\frac{sin \alpha}{sin \angle C - \alpha}.(\frac{a}{b} - cos \angle C) =? AK.\frac{AS}{BS}$$by $a - b.cos\angle C = c.cos\angle B$ and $AK = h_b = \frac{ac}{2R}$ and $2Rcos \angle B = BH$: $$\frac{sin \alpha}{sin \angle C - \alpha} =? \frac{BC}{BH} . \frac{AS}{BS}$$Now by ratio lemma , $\frac{sin \alpha}{sin \angle C - \alpha} = \frac{AS}{BS}. \frac{sin\angle SBC}{sin\angle SAC}$ And one can easily see that $\angle SBC = \angle HAS$, So $\frac{sin\angle SBC}{sin\angle SAC} = \frac{SH}{SC}$(by sine law in $(AHC)$). Finally it's enough to show that $\frac{BH}{BC}=\frac{SH}{SC}$ which has been already shown.

Let $CS$ meet the circle with diameter $AC$ at $Q\neq C$. Let $\omega$ be the circumcircle of triangle $BQC$. Redefine $P$ in such a way that the line $AC$ meet $\omega$ at $P$. We will prove that $\measuredangle APS=\measuredangle BAS$ instead. Denote the symbol $\measuredangle$ by directed angle. Let the line $AQ$ meet $\omega$ at $M$ and lines $AD$ and $PM$ meet at $N$. Since $\measuredangle AQC=\measuredangle CQM=90^\circ$, then $\measuredangle CPM=\measuredangle CBM=\measuredangle CQM=90^\circ$. Since $\measuredangle ASQ=\measuredangle ASC=\measuredangle AHC=\measuredangle DHF=\measuredangle DBA$ and $\measuredangle AQS=\measuredangle ADB$, then $\triangle AQS\sim\triangle ADB$. Therefore, $\measuredangle DAS=\measuredangle DAQ+\measuredangle QAS=\measuredangle DAQ+\measuredangle BAD=\measuredangle BAQ$. Since $\measuredangle QDN=\measuredangle QDA=\measuredangle QCA=\measuredangle QCP=\measuredangle QMP=\measuredangle QMN$, then $Q,D,N,M$ are concyclic. Next, we will show that $\triangle MBA\sim\triangle NSA$. Now, we have $\measuredangle NAS=\measuredangle DAS=\measuredangle BAQ=\measuredangle BAM$. It suffices to prove that $\frac{AM}{AN}=\frac{AB}{AS}$. Consider $$\frac{AM}{AN}=\frac{AD}{AQ}=\frac{AD}{AB}\cdot\frac{AB}{AQ}=\frac{AQ}{AS}\cdot\frac{AB}{AQ}=\frac{AB}{AS}$$Therfore, $\triangle MBA\sim\triangle NSA$. Consider $\measuredangle SCD=\measuredangle QCB=\measuredangle QMB=\measuredangle AMB=\measuredangle SNA=\measuredangle SND$, hence $S,C,N,D$ are concyclic. From $D,C,P,N$ are concyclic since $\measuredangle CDN=\measuredangle CPN=90^\circ$. Therefore, we have $S,C,P,N,D$ are conyclic. Finally, $\measuredangle APS=\measuredangle CPS=\measuredangle CDS=\measuredangle BDS=\measuredangle BAS$ as desired.

let $D=AH\cap BC$ , $E=BH\cap AC$ , $F=CH\cap AB$. Now $\angle APS$=$\angle BAS$=$\angle SDC$ then DSCP is cyclic. Now, if we define f as a geometric transformation to center c and radius $\sqrt{CD\cdot BC}$,Because $CD\cdot BC$ =$CH\cdot CF$ =$CE\cdot AC$ ,We have that: f(D)=B,f(B)=D,f(E)=A,f(A)=E,f(H)=F,f(F)=H,f(S)=$EF\cap(ADB)$,f(P)=$SB\cap AC$ , Now we have to prove that PD, CS and BE are congruent Note: For convenience in the figure, after the geometric transformation, we replaced H with D, A with D ,B with E(Look at the figure) According to the trigonometric form of Ceva's theory in $\triangle SEP$ , we must prove $$\frac {\sin \angle {SEB}}{\sin \angle {BEP}}\cdot \frac {\sin \angle {EPD}}{\sin \angle {DPS}}\cdot\frac {\sin \angle {CSP}}{\sin \angle {CSE}}=1$$ With the help of the law of sines in $\triangle CDP$and$\triangle BDP$: $$\frac {\sin \angle {EPD}}{\sin \angle {DPS}}=\frac{CD}{BD}\cdot\frac{PB}{PC}=\frac{\sin \angle{ABC}\cdot \sin \angle{90-ACB}}{\sin \angle{ACB}\cdot \sin \angle{90-ABC}}\cdot\frac {\sin \angle {ACB}}{\sin \angle {2ABC}}=\frac{\cos\angle{ACB}}{2(\cos\angle{ABC})^2} $$ With the help of the law of sines in $\triangle CBS$and$\triangle SEC$: $$\frac {\sin \angle {CSP}}{\sin \angle {CSE}}=\frac{BC}{CE}\cdot\frac{\sin\angle{2ABC}}{\sin\angle{ABC}}=\frac{1}{\cos\angle{ACB}}\cdot 2\cos\angle{ABC}$$ So we have: $$\frac {\sin \angle {SEB}}{\sin \angle {BEP}}\cdot \frac {\sin \angle {EPD}}{\sin \angle {DPS}}\cdot\frac {\sin \angle {CSP}}{\sin \angle {CSE}}=\cos\angle{ABC}\cdot\frac{\cos\angle{ACB}}{2(\cos\angle{ABC})^2}\cdot\frac{1}{\cos\angle{ACB}}\cdot 2\cos\angle{ABC}=1$$ So, PD,SC, and BE are collinear, and if we reverse f(The geometric transformation we defined), the problem is proved in its original form.

After inversion with center $A$ and radius $\sqrt{AH \cdot AD}$ we get the following problem: $\textbf{Problem.}$ In triangle $ABC$, $DEF$ is the orthic triangle. $S=DE \cap CF$. The line passing from $S$ and parallel to $AB$ intersects $AC$ at $P$. $I= BE \cap (PEF)$. Prove that $A,E,S,I$ are cyclic. Proof. Let $J$ be the midpoint of $HC$. Proving $AS \perp SI$ is enough. Brocard on the cyclic quadrilateral $HECD$ yields $AS \perp BJ$. Thus we will prove $IS \parallel BJ$. $(1)\triangle FAP \sim \triangle FHI \implies \frac{AP}{HI}=\frac{AF}{FH} \implies \frac{HI}{HB}=\frac{FH}{AF} \cdot \frac{AP}{HB}$ $(2) JH^2=JS \cdot JF \implies \frac{JS}{JH}=\frac{JH}{JF} \implies \frac{HS}{HJ}=\frac{HF}{JF}$ In the end, $\frac{HI}{HB}=\frac{HS}{HJ} \iff \frac{AF}{JF} =\frac{AP}{HB} \overset{S \text{ is the orthocenter of } ABJ}\iff \frac{FS}{FB} =\frac{AP}{HB} \iff \frac{AP}{FS} =\frac{HB}{FB} \iff \frac{AC}{CF} =\frac{HB}{FB} \leftarrow \triangle ACF \sim \triangle HBF $

How can we construct point P with ruler and compass? Thanks

DimPlak wrote: How can we construct point P with ruler and compass? Thanks I think $P$ lies on the circle tangent to $AB$ at $A$ and passing through $S$.

sami1618 wrote: DimPlak wrote: How can we construct point P with ruler and compass? Thanks I think $P$ lies on the circle tangent to $AB$ at $A$ and passing through $S$. Right! Thank you very much!