Lemma 1: $BL$ bisects $\angle ABC$ Let $BL \cap \odot ABC \equiv L'$. Consider an inversion through $B$ with radius $\sqrt{AB\cdot BC}$ with reflection through bisector of $\angle ABC$, and for clarity denote the inversion as a function $\psi$. Then $\psi (A)=C$ and reverse, $\psi (\Gamma) = \overline{AC}$ and reverse, and $\psi (\Omega)$ is a line through $\psi (L)$ parallel to $AC$. But we see that $\psi (L) \equiv L'$ reflected through angle bisector of $\angle B$ and so $\psi (L)$ had to be the midpoint of the arc $AC$ of $\Gamma$, and this implies that $BL$ is the actual angle bisector of $\angle B$. Lemma 2: $L$ is the center of $\odot MM_1M_2NN_1N_2$ Lemma 1 tells us $M_1 \in BC$ and $N_1 \in BA$. Also, by the reflection conditions we get $LM = LM_1 = LM_2 (1)$ and $LN = LN_1 = LN_2 (2)$. Also by lemma 1 we have $\angle NBL = \angle MBL \implies LN = LM$. Combining this with $(1)$ and $(2)$ gives the result Lemma 3: $B, N_1, M_1, K$ are concyclic $NN_2 \perp AC$ and $MM_2 \perp AC$ so $NN_2 \parallel MM_2$. Also, by the tangency condition we have $\angle NLC = \angle NBL = \angle MBL = \angle MLA$, and this along with reflection gives $\angle NLN_2 = \angle MLM_2$. Combining this with $NN_2 \parallel MM_2$ tells us $M, L , N_2$ are collinear, as well as $N, L, M_2$. Because L is the center of the circle given in lemma 2, we have $MN_1 \perp N_1N_2 \implies \angle KN_1B = \pi /2$ and $NM_1 \perp M_1M_2 \implies \angle KM_1B = \pi /2$, which give the result. For the next lemma keep in mind that $BK$ is the diameter of this circle Lemma 4: $L\in \odot B, N_1, M_1, K$ $M_1NN_1M$ is and isosceles trapezoid, so we have $M_1N_1 = MN = M_2N_2$, where the last equality follows from reflecting $MN$ through $L$. Also $M_1N_2 \parallel N_1M_2$ because $M_1N_2 \perp M_1M$ and $N_1M_2 \perp NN_1$, and obviously $MM_1 \parallel NN_1$. All of this tells us $N_1M_1N_2M_2$ is an isosceles trapezoid, so $KL \perp M_1N_2$ and $N_1M_2$ so $KL\parallel MM_1$. But we also have $BL\perp MM_1$ so $\angle KLB = \pi /2$, and this gives the result Final angle chase: We have $\angle KBN_1 =\angle KM_1N_1 =\angle M_2M_1N_1 =\angle M_2MN_1 \implies BK \parallel MM_2$, but $MM_2 \perp AC$ so $BK\perp AC$ $\blacksquare$

By Archimedes $BL$ bisects $\angle{ABC}$. Then $LN_{2}=LN_{1}=LN=LM=LM_{1}=LM_{2}$ Therefore $NM_{1}M_{2}MN_{1}N_{2}$ is cyclic. By Pascal on it we get that $B=NM_{1} \cap MN_{1};\; K=M_{1}M_{2} \cap N_{1}N_{2}$ and $X=M_{2}M \cap N_{2}N$ are collinear. But $X$ is a point at infinity for altitude from $B$. QED

Obviously, $L$ is the circumcenter of $\triangle MM_1M_2$ and $\triangle NN_1N_2$. By a well known homothety, we know that $BL$ is the angle bisector of $\angle ABC$ and since $BMLN$ is cyclic, we have $LM=LN$, implying that $MM_1M_2NN_1N_2$ are concyclic with center $L$. And by symmetry, we have $M,L,N_2$ and $N,L,M_2$ are collinear. Since $MN_1 = NM_1$, we have $\angle N_1N_2M = \angle KN_2L = \angle NM_2M_1 = \angle LM_2K \implies LKN_2M_2$ is cyclic. We have $\angle LKM_1= 180 - \angle LKM_2 = 180 - \angle LN_2M_2 = 180 - \angle MN_2M_2 = 180 - \angle MNM_2 = 180 - \angle MNL = 180 - \angle LBN \implies LBM_1K$ is cyclic. Since $N_1 \in (LBM_1)$, we have $LN_1BM_1K$ is cyclic. Finally, let $\angle MBL = x$, $\angle LBK = y$. We have $\angle M_2MB = 90 - x + 90 - y = 180 - x - y = 180 - \angle MBK \implies BK\parallel MM_2 \implies BK\perp AC$. $\square$

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