An application of the Casey’s theorem.

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.