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 2 of 6: Similar triangles give a length identity at H
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In plain words

A short angle chase using the two perpendiculars to BC (namely AE and SS') shows triangles AHD and BAH are similar, which converts the target length into a product of two known segments.

AH2=BH⋅HDAH^2=BH\cdot HD
Detailed analysis

Because S,O,S′S,O,S' are collinear (OO the circumcenter) and SS is an arc midpoint, SS′⊥BCSS'\perp BC; also AE⊥BCAE\perp BC by hypothesis, so AE∥SS′AE\parallel SS'. Equal arcs cut by these parallel chords give ∠EAS′=∠ABS\angle EAS'=\angle ABS. Writing φ=∠ABS\varphi=\angle ABS, this yields ∠DAH=φ=∠ABH\angle DAH=\varphi=\angle ABH, so triangles AHDAHD and BAHBAH share the angle at HH and satisfy ∠DAH=∠ABH\angle DAH=\angle ABH, hence △AHD∼△BAH\triangle AHD\sim\triangle BAH. Consequently AHBH=DHAH\dfrac{AH}{BH}=\dfrac{DH}{AH}, i.e. AH2=BH⋅HDAH^2=BH\cdot HD. Since B,D∈ωB,D\in\omega and HH lies on line BD (=BS)BD\,(=BS), the right-hand side is exactly the power of HH with respect to ω\omega.