Let $CX$ cross $B_1M$ at $N'$, $BX$ cross $C_1N$ at $M'$ and $BN'$ cross $CM'$ at $X'$ $\angle A_1N'C = \angle ABI$(because $A_1N' // AB$ and $N'C // BI$($BICX$ is paralelogram)) and $\angle A_1CN' = \angle CBI$($BICX$ is paralelogram). So $\angle A_1CN' = \angle A_1N'C$ and $A_1N' = A_1C$. Similarly, $A_1M' = BA_1$. Summing up, $BA_1 = A_1N' = A_1M = A_1C$ and $\angle CN'B = \angle BM'C = \angle 90^{\circ}$. So $XN'X'M'$ is concircle. X is orthocenter of triangle $BX'C$. $\angle MN'X = \angle MXN' = \angle BCX = \angle 90^{\circ} - \angle A_1BX' = \angle XX'N'$ so $XN'X'M'$ tangent to $A_1M, MN$. Similary $XN'X'M'$ tangent to $A_1N$. So $XN'X'M'$ is a $A_1$ -excircle of the triangle $A_1MN$. $\angle XN'X' = \angle IBX' = \angle 90^{\circ}$ so $BX'$ is ex-bisector of $\angle B$ Similarly, $CX'$ is ex-bisector of $\angle C$. So $X$ is excenters of the triangle $ABC$ and it is lie on $A_1$ -excircle of the triangle $A_1MN$ $\square$