I have just solved this problem (not very difficult). I will post my solution as soon as I have some free time. By the way, does anybody know how to transform constructions made in Cabri Geometry II Plus to a reasonable format like gif or jpg? :

Here is the solution. Let $o_{1}$ and $o_{2}$ be two circles tangent to rays $BA$ and $BC$ in points $X_{A},X_{C}$ and $Y_{A},Y_{C}$ respectively, so that $X_{A}A=Y_{C}C$. $o_{1}$ intersects line $AC$ in $E$ and $F$ (in order: $AEFC$) and $o_{2}$ intersects $AC$ in $E'$ and $F'$ (in order: $AF'E'C$). Denote $M$ the midpoint of $AC$. It is easy to verify that $M$ lies on the radical axis of $o_{1}$ and $o_{2}$. By the Power of a Point Theorem: $MF'\cdot ME'=MF\cdot ME$ Moreover, $AE\cdot AF=AX_{A}^{2}=CY_{C}^{2}=CE'\cdot CF'$ This implies, $AM^{2}+AM\cdot MF-ME\cdot AM-ME\cdot MF=(AM-ME)(AM+MF)=$ $=(CM-ME')(CM+MF')=CM^{2}+CM\cdot MF'-ME'\cdot CM-ME'\cdot MF'$ And finally we get: $MF-MF'=ME-ME'$ which combined with $\frac{MF}{MF'}=\frac{ME}{ME'}$ gives $ME=ME'$ and $MF=MF'$. So, if we now perform on circle $o_{2}$ symmetry with respect to $M$ it will be tangent to $DA$ and $DC$ and will pass through $E$ and $F$. This finishes the proof.