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 $$MA \cdot (CT-CZ) + MC \cdot (AX-AY) = AC\cdot (MR-MS).$$ 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.