Selection and Training Session

Let $\angle BAC=2\alpha$, $\angle ABC=2\beta$, $\angle BCA=2\gamma$. Since $AM=AN$, we obtain $\angle AMN=\angle ANM$. We have $\angle AMN=\angle ACN$ and $\angle MNC=\angle MAC$ (as inscribed angles). Further, $$\begin{array}{rl}\angle AKN& =\angle MAK+\angle AMK=\\ & =\angle MAC+\angle AMN=\\ & =\angle MNC+\angle ANM=\angle ANC.\end{array}$$ It follows that the triangles $AKN$ and $ANC$ are similar. Similarly, the triangles $ALM$ and $AMB$ are similar. Then $\frac{AN}{AC}=\frac{AK}{AN}$, so $A{I}^2=A{N}^2=AK\cdot AC$. In the same manner we see that $A{I}^2=AL\cdot AB$, so $AK\cdot AC=AL\cdot AB$, which shows that $L$, $K$, $C$, $B$ are concyclic. Then $\angle AKL=\angle ABC=2\beta$. Moreover, since $AI:AK=AC:AI$, it follows that the triangles $AIK$ and $ACI$ are similar. Then $\angle AIK=\angle ICA=0.5\angle BCA=\gamma$. Since $\angle IKC=\angle IAK+\angle AIK=\alpha +\gamma$, we have $$\begin{array}{rl}\angle IKL& =180^{\circ }-\angle AKL-\angle IKC=180^{\circ }-2\beta -\alpha -\gamma =\\ & =180^{\circ }-2(\alpha +\beta +\gamma )+\alpha +\gamma =[\alpha +\beta +\gamma =90^{\circ }]=\alpha +\gamma =\angle IKC.\end{array}$$ Similarly, $\angle KLI=\angle ILB$. Therefore, all bisectors of the quadrilateral $LBCK$ intersect at the same point $I$. Hence $LBCK$ is a circumscribed quadrilateral. Thus $MN$ touches the inscribed circle of the triangle $ABC$.