Let $ABC$ be an acute-angled triangle with $AB<AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E\ne A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P\ne B$.

Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.

Step 1: $S$ is the midpoint of arc $BC$ containing $A$, so $SA=SB=SC$ (no — $S$ is the arc midpoint, so $SB=SC$ and $AS$ is the angle bisector of $\angle BAC$).

Step 2: $AE\perp BC$ with $E$ on $\Omega$. $D=AE\cap BS$. The line through $D$ parallel to $BC$ meets $BE$ at $L$.

Step 3: Since $DL\parallel BC$ and $D\in BS$: triangle $BDL$ has specific angle relations with $△BEF$ (where $F$ is the foot). The circumcircle $\omega$ of $BDL$ has computable angles.

Step 4: $P$ is the second intersection of $\omega$ and $\Omega$. The tangent to $\omega$ at $P$ makes angle $\angle BPD$ with $PB$ (by the tangent-chord angle). Using inscribed angles in both circles, chase the angle to show the tangent meets $BS$ at a point on the angle bisector from $A$.

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