$H$ is the orthocenter of $\triangle ABC$,the circle with center $H$ passes through $A$,and it intersects with $AC,AB$ at two other points $D,E$.The orthocenter of $\triangle ADE$ is $H'$,line $AH'$ intersects with $DE$ at point $F$.Point $P$ is inside the quadrilateral $BCDE$,so that $\triangle PDE\sim\triangle PBC$.Let point $K$ be the intersection of line $HH'$ and line $PF$.Prove that $A,H,P,K$ lie on one circle.
Problem
Source: 2022 China Southeast Grade 10 P6
Tags: geometry
Mahdi_Mashayekhi
03.08.2022 22:05
Let $S$ be intersection of $BD$ and $CE$. Since $\triangle PDE\sim\triangle PBC$ we have $\triangle PDB\sim\triangle PEC$ so $PDES$ and $PCBS$ are cyclic. Let $BC$ meet $DE$ at $M$. Let $AH$ meet $BC$ at $G$ and $AF$ meet $ADE$ at $T$. Let $HH'$ meet $DE$ at $R$. Claim $: BEHDC$ s cyclic. Proof $:$ Note that $\angle HEA = \angle HAE = \angle 90 - \angle B = \angle HCB \implies HCBE$ is cyclic. with same approach we prove that $HBCD$ is cyclic. Claim $: H',H,S$ are collinear. Proof $:$ Note that $HH'$ is Euler line of $ADE$ and $BEDC$ is cyclic so intersections of diagonals of $BEDC$ which is $S$ lies on $HH'$. Claim $: \angle APM = \angle 90$. Proof $:$ $P$ is Miquel point by how we redefined it. Claim $: PF$ and $HH'$ meet at $DPE$. Proof $:$ Note that we need to prove $\angle DPF = \angle DSH$. Note that $P$ sends $DE$ to $BC$ so $\angle DPF = \angle BPG = \angle MPG - \angle MPB = \angle MPG - \angle SPB = \angle MPG - \angle SDE$. Note that Since $\angle APM = \angle 90 = \angle AFM = \angle AGM$ then $AFPGM$ is cyclic so $\angle MPG - \angle SDE = \angle SAG - \angle SDE = \angle MAH - \angle SDE$ and since $R$ is intersection of Euler line with $DE$ then simply $AHRM$ is cyclic so $\angle MAH - \angle SDE = \angle MRS - \angle SDE = \angle MRS - \angle SDM = \angle DSO = \angle DSH$ so $\angle DPF = \angle DSH$ as wanted. $\angle APK = \angle APF = \angle FMA = \angle RMA = \angle RHG = \angle AHK \implies KAHP$ is cyclic so we prove that $KAHP$ is cyclic (It can easily be proved that in fact $T$ lies on this circle as well). We're Done (yes I know we didn't need to prove $KDPE$ is cyclic but anyway...)
parmenides51
06.04.2024 10:32
$H$ is the orthocenter of $\triangle ABC$,the circle with center $H$ passes through $A$,and it intersects with $AC,AB$ at two other points $D,E$.The orthocenter of $\triangle ADE$ is $H'$,line $AH'$ intersects with $DE$ at point $F$.Point $P$ is inside the quadrilateral $BCDE$,so that $\triangle PDE\sim\triangle PBC$.Let point $K$ be the intersection of line $HH'$ and line $PF$.Prove that $A,H,P,K$ lie on one circle.