An application of the Casey's theorem. 

Let $AB$ and $CD$ be two chords of a circle $\Phi$, which intersects in the point $E$.
Circles $\Phi_1$ and $\Phi_2$ are placed inside $\Phi$ such that $\Phi_1$ touches to segments $AE$ and $DE$ and to the circle $\Phi$ and $\Phi_2$ touches to segments $BE$ and $CE$ and to the circle $\Phi$.
Let $l$ be a common external tangent to $\Phi_1$ and to $\Phi_2$ such that $l$ intersects segments $AE$ and $CE$.


Prove that $l||AC.$

I solved this problem $25$ years ago by using the Casey's theorem, but I forgot, how I made it.
I am looking for a proof by the Casey's theorem only.
This problem also $4.7.29$ from a book of A.Akopyan "Geometry in Figures".
We can make here the following things.
We can use  Casey for  "quadrilaterals" $A\Phi_2D\Phi_1$, $C\Phi_2D\Phi_1$ or even for $AC\Phi_2\Phi_1$,
but I don't see how it may help.
About the Casey's theorem see here:  https://en.wikipedia.org/wiki/Casey%27s_theorem
Thank you!
 A: 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.
A: COMMENT:

If points A, C, L and K can be considered as circles with zero diameter then we may write:
$$AC\times RS+AM\times CN=CU\times AT$$
$$LN \times KM+ RS \times KL =LU\times KT$$
Comparing thses relations we may conclude that AC is transformed in parallel order to KL.
Can this help you?
A: I got something, but it still does not give a proof.
Let the external tangent be touched to $\Phi_1$ and $\Phi_2$ at points $M$ and $N$ respectively,
$AB$ be touched to $\Phi_1$ and $\Phi_2$ at points $T$ and $H$ respectively and
$DC$ be touched to $\Phi_1$ and $\Phi_2$ at points $F$ and $S$ respectively.
Thus, by Casey for $AC\Phi_2\Phi_1$ we obtain:
$$AC\cdot MN+AT\cdot CS=AH\cdot CF$$ or
$$AC\cdot MN+AT\cdot CS=(AT+TH)(CS+SF)$$ and since $TH=SF$, we obtain:
$$AC\cdot MN=TH(AT+CS+TH)$$ or
$$AC\cdot MN=TH(AT+TE+CS+SE)$$ or
$$AC\cdot MN=TH(AE+CE)$$ or
$$\frac{AC}{AE+CE}=\frac{TH}{MN}.$$
Now, let $\Phi$ be changed, but lines $AB$ and $CD$ and $\Phi_1$ and $\Phi_2$ be not changed.
Let $A'$ and $C'$ be such point on new $\Phi'$.
Thus, $$\frac{AC}{AE+CE}=\frac{A'C'}{A'E+C'E}.$$
Now, let $AE=a$, $CE=b$, $A'E=x$, $C'E=y$ and $\measuredangle AEC=\alpha.$
Thus, $$\frac{a^2+b^2-2ab\cos\alpha}{(a+b)^2}=\frac{x^2+y^2-2xy\cos\alpha}{(x+y)^2}$$ or
$$(a^2+b^2)(x^2+y^2+2xy)-(x^2+y^2)(a^2+b^2+2ab)=2\cos\alpha(ab(x+y)^2-xy(a+b)^2)$$ or
$$(xy(a^2+b^2)-ab(x^2+y^2))(1+\cos\alpha)=0$$ or
$$xy(a^2+b^2)-ab(x^2+y^2)=0$$ or
$$(ay-bx)(ax-by)=0.$$
Now, $ay=bx$ gives $AC||A'C'$ and from here easy to complete the proof.
$ax=by$ gives $A'C'||BD$.
It seems that it's impossible, but how to prove it?
