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A lot of symmetry

Source: Southern Summer School, gr. 12

July 9, 2017
geometrygeometric transformationreflectionsymmetry

Problem Statement

Let ω\omega be a circle with center OO and a non-diameter chord BCBC of ω\omega. A point AA varies on ω\omega such that ∠BAC<90∘\angle BAC<90^{\circ}. Let SS be the reflection of OO through BCBC. Let TT be a point on OSOS such that the bisector of ∠BAC\angle BAC also bisects ∠TAS\angle TAS. 1. Prove that TB=TC=TOTB=TC=TO. 2. TB,TCTB, TC cut ω\omega the second times at points E,FE, F, respectively. AE,AFAE, AF cut BCBC at M,NM, N, respectively. Let SMSM intersects the tangent line at CC of ω\omega at XX, SNSN intersects the tangent line at BB of ω\omega at YY. Prove that the bisector of ∠BAC\angle BAC also bisects ∠XAY\angle XAY.
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