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 6 of 6: Conclude the tangency
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In plain words

Once HP equals HA, its square matches the power of H with respect to omega computed earlier, and a segment from an external point whose square equals the power of that point with respect to a circle through its far endpoint must be tangent there.

HP2=pow⁡(H,ω)  ⟹  HP tangent to ω at PHP^2=\operatorname{pow}(H,\omega)\implies HP\text{ tangent to }\omega\text{ at }P
Detailed analysis

By Step 5 and Step 2, HP2=HA2=BH⋅HD=pow⁡(H,ω)HP^2=HA^2=BH\cdot HD=\operatorname{pow}(H,\omega). Since P∈ωP\in\omega, this is precisely the condition for line HPHP to be tangent to ω\omega at PP (the converse of the power-of-a-point theorem). Therefore the tangent to ω\omega at PP passes through HH, which lies on line BSBS and on the internal bisector AS′AS' of ∠BAC\angle BAC, completing the proof.