, 2013

Let $D$ be the midpoint of arc $BC$, $N$ the midpoint of $BC$, $X$ the orthogonal projection of $P$ onto $AC$. Then points $P, X, K, L, N, C$ are concyclic (Fig. 7).

Fig. 7

We have $\angle PNX = \angle PCA = \angle PDA$ therefore $XN \parallel KL$. It follows that $LN = KX$, and $\angle LCN = \angle KCA$. Obviously $\angle BAK = \angle LAC$, so $K, L$ are isogonal conjugates with respect to triangle $ABC$. Thus following equalities hold: $\angle MBC = \angle CBL = \angle KBA$ and $\angle BCM = \angle LCB = \angle ACK$. This implies that $A, M$ are isogonal conjugates with respect to triangle $KBC$. Using this we get $\angle BNM = \angle LNB = \angle LKC = \angle BKM$ therefore points $B, M, N, K$ are concyclic.

In the first solution, it is proven that points $K, L$ are isogonal conjugates with respect to $ABC$. Thus angle bisectors of $\angle CBA$ and $\angle BCA$ are angle bisectors of $\angle LBK$ and $\angle KCL$, respectively. From angle bisector theorem we deduce that $BK/BL = KC/LC$. This means that points $K, L$ lie on Apollonian circle $\omega$ of points $B, C$. Since $\omega$ is self-symmetric with respect to line $BC$, we get that $M$ lies on $\omega$ (Fig. 8).

Fig. 8

Let $\omega$ intersect segment $BC$ at point $Y$. Then $KY$ is angle bisector of both angles $\angle MKL$ (because $Y$ is the midpoint of arc $LM$) and $\angle BKC$ (because $K$ lies on Apollonian circle). Therefore $\angle BKM = \angle LKC$. We end proof as we did in the first solution.