Let $\triangle ABC$ be an acute angle triangle and $\omega$ be its circumcircle. Let $N$ be a point on arc $AC$ not containing $B$ and $S$ be a point on line $AB$. The line tangent to $\omega$ at $N$ intersects $BC$ at $T$, $NS$ intersects $\omega$ at $K$. Assume that $\angle NTC = \angle KSB$. Prove that $CK\parallel AN \parallel TS$.
Problem
Source: BdMO 2023 Secondary National P6
Tags: geometry, cyclic quadrilateral, Angle Chasing
seojun8978
12.02.2023 17:19
(N.T.B.S concyclic) for obvious reasons. So $\angle ABC = \angle SNX$ let X be the point on the tangent of N and also on the other side of the point T. So, $\angle KCN=\angle SNX=\angle ABC$ therefore, $KN=AC$ So, $AN$, $CK$ parallel. Also, $\angle CKN=\angle CBN=\angle TBN=\angle TSN$ therefore, $CK$, $TS$ parallel.
Seungjun_Lee
12.02.2023 17:20
Trivial that $S,B,T,N$ is concyclic $SN \cap BC = X$ $XS \times XN = XB \times XT$ and $XK \times XN = XB \times XC$ Therefore, $KC \parallel ST$ Let $BX \cap TN = Y$ $YA \times YB = YN^2$ $YS \times YB = YN \times YT$ Therefore, $AN \parallel ST$ Hence,, $CK \parallel AN \parallel TS$
wasikgcrushedbi
29.02.2024 16:54
Let $KC \cap AB=P$. Then $\triangle PSK$ and $\triangle CNT$ are similar since $\angle CNT=\angle PKS$. So, $\angle BAN=\angle NCT=\angle BPC \implies KC||AN$. Then $\angle NKC=\angle NBC=\angle NST$ (since $BSNT$ concyclic) gives $KC||ST$