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Let $\triangle ABC$ be an acute 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 $N$ interesects $BC$ at $T$, $NS$ intersects $\omega$ at $K$.

Assume that $\angle NTC=\angle KSB$. Prove that $CK\parallel AN\parallel TS$.


Proof Based Problems  


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Solution

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Since $\angle KSB=\angle NTC = \angle NTB$ , hence $\angle NSB + \angle NTB = (180^\circ-\angle KSB ) +\angle NTB=180^\circ$.

Therefore, $B,S,N,T$ are concyclic. Now, $\angle KCB=\angle KNB=\angle SNB=\angle STB$.

So, $TS\parallel CK$.

Since $NT$ is tangent to $\omega$, hence $\angle BAN=\angle BNT$.

Also $\angle BST=\angle BNT=\angle BAN$.

$\therefore AN\parallel TS$.

Since $TS\parallel CK$ and $TS\parallel AN$, hence $CK\parallel AN\parallel TS$.

This is a proof based problem added for learning purposes and does not accept submissions.

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