$tan \angle MB_1B=\frac{AB_1 \cdot \frac{BA_1}{BC}}{h_B \cdot \frac{CA_1}{BC}}= \frac{AB_1 \cdot BA_1}{h_B \cdot CA_1}$ and $tan \angle KA_1A=\frac{BA_1 \cdot \frac{AB_1}{AC}}{h_A \cdot \frac{B_1C}{AC}} = \frac{BA_1 \cdot AB_1}{h_A \cdot B_1C}$ so to prove$\angle AA_1K=\angle BB_1M \Leftrightarrow$ to prove $\frac{h_B}{h_A}=\frac{B_1C}{A_1C}$ It comes from a pair of familiar similar triangle $\triangle AA_1C \sim \triangle BB_1C$

The height AA1 + B1K//BC =>the angle A1EB1=90 . For the same reason,if we make D and E the intersection of A1M and BB1,AA1 and B1K,the angle A1DB1=90. So the points A1,D,E,B1 are on the same circle => the angle DA1E=the angle DB1E. B1FA1C is a parallelogram, and the angle A1AC = the angle CBB1.So: MF/FK=AC/BC=(A1C/sin( A1AC ))/(B1C/sin( CBB1 ))=(B1F/sin( A1AC ))/(A1F/sin( CBB1 ))=B1F/A1F. So MF*FA1=KF*FB1 => the points M,K,A1,B1 are on the same circle => the angle MA1K=the angle MB1K. So the angle MA1K + the angle MA1E =the angle MB1K + the angle KB1D => the angle AA1K= the angle BB1M.(there is another situation:There is no intersection between line MA1 and line KB1,this time we need turn "+" into "-") I don't konw how to send something like #2...

Since $B_1K \parallel BC$, hence $\angle A_1BB_1=\angle CBB_1=\angle BB_1K$. Since $A_1M \parallel AC$, hence $\angle B_1AA_1=\angle CAA_1=\angle AA_1M$. We know that $AB_1A_1B$ siklik, so we have $\angle B_1AA_1=\angle A_1BB_1$, so now we have $\angle AA_1M=\angle BB_1K$. Observe that $$ \angle BMA_1=180^\circ-\angle MBA_1-\angle MA_1B=180^\circ-\angle CB_1A_1-\angle B_1CA_1=\angle CA_1B_1=\angle A_1B_1K$$So we have $A_1B_1KM$ cyclic. Hence $\angle MA_1K=\angle MB_1K$. So its proved that $\angle AA_1K=\angle BB_1M$.