Bài toán mở rộng về đường tròn chín điểm ( Tiếng Anh)

Let $ABC$ be a triangle with circumcircle $(O)$. $(K)$ is a circle passing through $B,C$. $(K)$ cuts $CA,AB$ again at $E,F$. $BE$ cuts $CF$ at $H_K$.

a) Prove that $H_KK$ and $AO$ intersect on $(O)$.

b) $O_K$ is isogonal conjugate of $H_K$ with respect to triangle $ABC$. Prove that $O_K$ lies on $OK$.

c) Let $L,N$ be the points on $CA,AB$, resp such that $O_KL\parallel BE, O_KN\parallel CF$. Prove that $LN\parallel BC$.

d) The line passing through $N$ parallel to $BE$ cuts the line passing through $L$ parallel to $CF$ at $P$. Prove that $P$ lies on $AH_K$.

e) $Q,R$ lie on $BE,CF$, resp such that $PQ\parallel AB,PR\parallel AC$. Prove that $QR\parallel BC$.

f) Prove that $NQ,LR$ and $AH_K$ are concurrent.

g) $D$ is projection of $K$ on $AH_K$. Prove that $DK,EF,BC$ are concurrent.

h) Prove that $KN\perp BE, KL\perp CF$.

i) Prove that nine points $D,E,F;P,Q,R;K,L,N$ lie on a circle $(N_K)$.

j) Prove that $N_K$ is midpoint of $PK$ and $KN_K$ is parallel to $AO$.

k) Prove that $H_K,N_K,O$ are collinear.

When $K \equiv M$ midpoint of $BC$, we get all properities of Nine-point circle.

Solution:

a, g) Let $S \equiv EF \cap BC.$ Then $AS$ is the polar of $H_K$ WRT $(K)$ and $AH_K$ is the polar of $S$ WRT $(K)$ $\Longrightarrow$ $KH_K$ is perpendicular to $AS$ through $H$ and $AH_K$ is perpendicular to $KS$ through $D.$ Hence $SH \cdot SA=SD \cdot SK=SB \cdot SC$ $\Longrightarrow$ $H \in (O).$ Since $\angle AHH_K=90^{\circ},$ then $KH_K$ and $AO$ meet on $(O).$

b, c) $\angle O_KBC=\angle H_KBF=\angle H_KCE=\angle O_KCB$ $\Longrightarrow$ $O_K$ is on perpendicular bisector $OK$ of $\overline{BC}.$ $\angle BFC=\angle BNO_K=\angle BKO_K$ (mod 180) $\Longrightarrow$ $N,B,K,O_K$ are concyclic $\Longrightarrow$ $\angle BNK=\angle BO_KK.$ But $\angle BO_KK=90^{\circ}-\angle FBE=\angle FEK$ $\Longrightarrow$ $\angle BNK=\angle FEK,$ i.e. $N$ lies on circumcircle $(N_K)$ of $DKEF.$ Similarly, $L \in (N_K).$ Thus, $LN$ is antiparallel to $EF$ WRT $AE,AF$ $\Longrightarrow$ $LN \parallel BC.$

d, f)$\triangle PLN$ and $\triangle H_KCB,$ with parallel sides, are homothetic with center $A$ $\Longrightarrow$ $A,P,H_K$ are collinear. Likewise, $\triangle ANL$ and $\triangle PQR,$ with parallel sides, are homothetic with center $AP \cap NQ \cap LR,$ i.e. $AH_K,NQ,LR$ concur.

i) $\angle NKL=\angle NBO_K+\angle LCO_K=CBH_K+\angle BCH_K=\angle NPL$ (mod 180) $\Longrightarrow$ $P \in (N_K).$ Further, $P$ is the midpoint of the arc $EF$ of $(N_K),$ because $\angle PFK=\angle PDK=90^{\circ},$ i.e. $KP$ is perpendicular bisector of  $\overline{EF}.$ Now, since $\angle PQE=\angle FBE=\angle PKE,$ it follows that $Q \in (N_K).$ Similarly, $R \in (N_K).$

h, j) $D,E,F,P,Q,R,K,L,N$ lie then on a circle $(N_K)$ with diameter $KP$ perpendicular to $EF,$ i.e. $KP \parallel AO.$ Thus, $KN$ is perpendicular to $PN \parallel BE$ and $KL$ is perpendicular to $PL \parallel CF.$

e, k) $\triangle PQR \cup (N_K)$ and  $\triangle ABC \cup (O)$ are homothetic with center $H_K \equiv AP \cap BQ \cap CR,$ thus $QR \parallel BC$ and $H_K,O,N_K$ are collinear.