Cute, first if $AB=BC$ then $\Gamma_1$ and $\Gamma_2$ are symetric w.r.t. the perpendicular bisector of $BC$ so $E,F$ are symetric w.r.t. the perpendicular bisector of $BC$ which means $EF \parallel BC$. Now WLOG $AB>BC$ and let $L$ the center of $\Gamma_1$ and $M$ the center of $\Gamma_2$ also let $\Delta$ intersect $BC$ at $D$, now by easy angle chasing $\angle ADB=\angle ABC-\angle ACB$, now its clear that $D,L,M$ are colinear and also $\angle AML=\frac{\angle ABC-(\angle ABC-\angle ACB)}{2}=\frac{\angle ACB}{2}=\angle DAL$ which means that $\Delta$ is tangent to $(ALM)$, now let $AB \cap (ALM)=A'$ and $AC \cap (ALM)=A''$ then by reims on $\Gamma, (ALMA'A'')$ we get $BC \parallel A'A''$ now $\angle LA'A=\angle DAL=\angle LAA'$ which means that $AL=LA'$ so $E$ is midpoint of $AA'$ and in a similar way u get $F$ midpoint of $AA''$ hence by midbase we get that $BC \parallel A'A'' \parallel EF$ thus we are done

Let $D$ be $\Delta \cap BC$ , It suffices to show $\triangle AEF \sim \triangle ABC \newline$ Claim 1: $\frac{AE}{AB} = \frac{AF}{AC} \newline$ Using the formulae for points of tangency for $\Gamma_1$ and $\Gamma_2$ and simplifying we obtain the question is equivalent to showing $AB(AD-CD) = AC(BD-AD) \newline$ Now note $\triangle ADB \sim \triangle CDA$. Hence the question simplifies to $AB*CD = AC*AD$ which holds due to the aforementioned similarity $\newline$ From here $\triangle AEF \sim \triangle ABC$ follows. QED