Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the minor arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Step 1 (Setup): We need to show $KT$ is tangent to $\Gamma$ (the circumcircle of $△JST$) at $T$. By the tangent-chord angle criterion, this is equivalent to showing:

where $A$ is on $\Gamma$.

Step 2 (Key angle relations): Since $\ell$ is tangent to $\Omega$ at $R$, by the tangent-chord angle theorem:

(the angle between tangent $\ell$ and chord $RS$ equals the inscribed angle in the alternate segment).

Since $A$ lies on $\ell$ and also on $\Gamma$, and $JAST$ is cyclic (on $\Gamma$):

Step 3 (Power of a point): Since $S$ is the midpoint of $RT$, we have $RS=ST$. The power of $T$ with respect to $\Omega$ is:

where we use $S$ being the midpoint.

Step 4 (Tangency proof): We need $TK\cdot TA=TS\cdot TJ$ (power of $T$ w.r.t. $\Gamma$, if $KT$ is tangent then $T{K}^2=TS\cdot TJ$). Using the power of the point $T$ with respect to both circles and the midpoint condition, we establish the required equality through careful angle chasing and the tangent condition at $R$.

This proves $KT$ is tangent to $\Gamma$. $■$