MathLabs

Problem 4

Let RR and SS be different points on a circle Ω\Omega such that RSRS is not a diameter. Let ℓ\ell be the tangent line to Ω\Omega at RR. Point TT is such that SS is the midpoint of RTRT. Point JJ is chosen on the shorter arc RSRS of Ω\Omega so that the circumcircle Γ\Gamma of triangle JSTJST intersects ℓ\ell at two distinct points. Let AA be the common point of Γ\Gamma and ℓ\ell that is closer to RR. Line AJAJ meets Ω\Omega again at KK. Prove that line KTKT is tangent to Γ\Gamma.
Step 2 of 6: Obtain the key parallelism
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∠RKA=∠RKJ=∠RSJ=∠TSJ=∠TAJ=∠TAK,RK∥AT\angle RKA=\angle RKJ=\angle RSJ=\angle TSJ=\angle TAJ=\angle TAK,\qquad RK\parallel AT
Detailed analysis

Use directed angles modulo 180 degrees. The first equality uses A, J, K collinear; the second uses R, S, J, K on the same circle; the third uses R, S, T collinear; the fourth uses S, T, A, J on the same circle; and the last again uses A, J, K collinear. Hence RK is parallel to AT.