Let $M$ be the midpoint of the arc $KL$ containing points $A$ and $C$. Denote by $R, S, X, Y, Z, T, U, V$ points of tangency as in the picture.
Our strategy is to prove that $MA=MC$ because then $M$ is the midpoint of arc $AC$ as well which leads to $AC parallel KL$.
Using Casey’s theorem for $M$, $K$, $Phi_1$, and $L$ we obtain
$$MK cdot LU + ML cdot KU = MR cdot KL.$$
Since $KU+LU=KL$ and $MK=ML$, it follows that $MR=MK$.
Analogously, using Casey’s theorem for $M$, $K$, $Phi_2$, and $L$ we obtain $MK=MS$.
Casey’s theorem for $A$, $M$, $C$, and $Phi_1$ gives
$$MA cdot CT + MC cdot AX = AC cdot MR.$$
Similarly, Casey’s theorem for $A$, $M$, $C$, and $Phi_2$ yields
$$MA cdot CZ + MC cdot AY = AC cdot MS.$$
Subtracting the two equalities we obtain
$$MA cdot (CT-CZ) + MC cdot (AX-AY) = ACcdot (MR-MS).$$
Since $CT-CZ=ZT$, $AY-AX=XY$, and $MR=MK=MS$, we obtain
$$MA cdot ZT + MC cdot (-XY) = 0.$$
But $ZT=XY$, hence $MA-MC=0$. Thus $MA=MC$ and we are done.