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 6 of 6: Finish by the tangent–chord theorem
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∠KTA=∠TSA⟹KT is tangent to Γ at T\angle KTA=\angle TSA\Longrightarrow KT\text{ is tangent to }\Gamma\text{ at }T
Detailed analysis

The angle between KT and the chord TA equals the inscribed angle subtending the same chord TA in Gamma. By the converse of the tangent–chord theorem, KT is tangent to Gamma at T, as required.