BMOSL 2008 G5 wrote: The circle $k_a$ touches the extensions of sides $AB$ and $BC$, as well as the circumscribed circle of the triangle $ABC$ (from the outside). We denote the intersection of $k_a$ with the circumscribed circle of the triangle $ABC$ by $A'$. Analogously, we define points $B'$ and $C'$. Prove that the lines $AA',BB'$ and $CC'$ intersect in one point. Hmm this is too well known. A $\sqrt{AB\cdot AC},\sqrt{BC\cdot BA},\sqrt{CA\cdot CB}$ Inversion with a flip around the respective angular Bisectors swaps $\{\mathcal K_a,\mathcal K_b,\mathcal K_c\}$ swaps to the Incircle of $\triangle ABC$.Now taking only the Isogonal Lines into Consideration we get that $\{AA',BB',CC'\}$ concur at the Isogonal Conjugate of the Gregonne Point $(X_{55})$ of $\triangle ABC$. $\blacksquare$