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 3 of 6: A diameter appears from the second solution circle
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In plain words

Chasing angles around the two circles shows the segment EF (with F the second meeting point of line PD with Omega) is parallel to BC, and since AE is already perpendicular to BC that forces AF to be a diameter.

AF is a diameter of Ω,F=PD∩ΩAF\text{ is a diameter of }\Omega,\quad F=PD\cap\Omega
Detailed analysis

Let F≠AF\ne A be the second intersection of line PDPD with Ω\Omega. Using ω\omega and Ω\Omega, ∠CBE=∠DLE=∠DPB=∠FCB\angle CBE=\angle DLE=\angle DPB=\angle FCB, so EF∥BCEF\parallel BC. Since AE⊥BCAE\perp BC, also AE⊥EFAE\perp EF, so ∠AEF=90∘\angle AEF=90^\circ is inscribed in Ω\Omega and subtends a diameter: AFAF is a diameter of Ω\Omega.