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Let $P$ be the intersection of the lines $p$ and $KL$. Denote $\angle TBC=x$. Line $BT$ is a transversal of the parallel lines $p$ and $q$, which implies that $\angle BTA=\angle TBC=x$. The quadrilateral $BCLK$ is cyclic, hence $\angle CLK=180^{\circ }-\angle KBC=180^{\circ }-x$, implying $\angle KLT=x$. Triangles $PKT$ and $PTL$ are similar because they share an angle at $P$ and $\angle KTP=\angle PLT=x$. This implies $$\frac{|PK|}{|PT|}=\frac{|PT|}{|PL|},\text{i.e.}|PK|\cdot |PL|=|PT{|}^2.$$ Since the power of the point $P$ with respect to the circle $k$ is $|PK|\cdot |PL|=|PA{|}^2$, we can conclude that $|PA|=|PT|$, i.e. the point $P$ is the midpoint of $\overline{AT}$.