MathLabs

Problem 2

Let ABCABC be an acute-angled triangle with AB<ACAB<AC. Let Ω\Omega be the circumcircle of ABCABC. Let SS be the midpoint of the arc CBCB of Ω\Omega containing AA. The perpendicular from AA to BCBC meets BSBS at DD and meets Ω\Omega again at E≠AE\neq A. The line through DD parallel to BCBC meets line BEBE at LL. Denote the circumcircle of triangle BDLBDL by ω\omega. Let ω\omega meet Ω\Omega again at P≠BP\neq B. Prove that the line tangent to ω\omega at PP meets line BSBS on the internal angle bisector of ∠BAC\angle BAC.
Step 1 of 6: Name the target point on the bisector
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In plain words

The antipode of the arc midpoint S is the classical arc midpoint that lies on the internal bisector from A, so intersecting it with line BS names exactly the point we must put the tangent line through.

H=BS∩AS′,S′=antipode of SH=BS\cap AS',\qquad S'=\text{antipode of }S
Detailed analysis

Since SS is the midpoint of arc CBCB containing AA, its antipode S′S' with respect to Ω\Omega is the midpoint of arc BCBC not containing AA, which is the classical point where the internal bisector of ∠BAC\angle BAC meets Ω\Omega. Let H=BS∩AS′H=BS\cap AS'. Proving the statement now amounts to showing that HH lies on the tangent to ω\omega at PP, because HH already lies on both BSBS and the bisector AS′AS'.