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 4 of 6: Match two ratios along the same transversal
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In plain words

Comparing the midline segment from the circumcenter with the segment AD along the two lines perpendicular to BC produces equal ratios, which is exactly the condition for OH to be parallel to the chord PD.

OH∥DQ,Q=PD∩SS′OH\parallel DQ,\qquad Q=PD\cap SS'
Detailed analysis

Let Q=PD∩SS′Q=PD\cap SS' and let OO be the center of Ω\Omega. Since OO is the midpoint of the diameter AFAF found in Step 3 and QQ lies on chord PFPF, a similar-triangles argument along the parallel lines SS′∥AESS'\parallel AE gives OQ=12ADOQ=\tfrac12AD. Combining this with SO=12SS′SO=\tfrac12SS' gives SOOQ=SS′AD=SHHD\dfrac{SO}{OQ}=\dfrac{SS'}{AD}=\dfrac{SH}{HD} (the last equality from the similar triangles of Step 2 along transversal BSBS), and this proportion along the transversal lines S,O,HS,O,H and S,Q,DS,Q,D forces OH∥DQOH\parallel DQ, i.e. OH∥PDOH\parallel PD.