Let $AC$ intersect $A_1A_2$ in F', and let the incenter of $CC_1C_2$ be tangent to $C_1C_2$ at $F''$. Looking at the triangle $AA_1A_2$, $E$ is the antipode of $F$-the tancengy point in the incircle of the triangle, hence is well known that $F'$ will be the tancency point of the excircle in the respective triangle, hence $A1F=F'A_2$. By the homothety sending the incircle of the triangle $CC_1C_2$ to the excircle of the respective triangle(which happens to be the incircle of a quadrilateral $ABCD$), we get $C,F'',E$ collinear, but $CE$ intersects $C_1C_2$ in $F'$, hence $F'=F''$. As $A_1F=F'A_2$ and $C_1F=F'C_2$, the conclusion follows.