Nice Problem Let $B_1A_2 \cap (I)=D$ and $B_1C_2 \cap (I)=E$, then $A_1I$ is the perpendicular bisector of $EB_1$ $\implies$ $E-A_1-A_2$, similarly, $C_2-C_1-D$ And since, $\angle EB_1D=90^{\circ} \implies E-I-D$. Obviously, $A,C$ are the centers of $\odot (B_1C_1C_2)$ $ ,$ $ \odot (B_1A_1A_2)$. Let $A_1A_2 \cap C_1C_2=F$ $$\angle C_1EF=\angle C_1B_1A_1=180^{\circ}-(\angle AB_1C_1+\angle CB_1A_1)=45^{\circ}$$Hence, $B$ is the center of $\odot (C_1FA_1)$. Now, Let $M$ be midpoint of $AC$ $$\angle ABM=\angle BAC=2\angle C_2B_1C_1=2\angle C_1A_1E=\angle ABF$$and hence, $B-F-M$

parmenides51 wrote: The incircle of right-angled triangle $ABC$ ($\angle B = 90^o$) touches $AB,BC,CA$ at points $C_1,A_1,B_1$ respectively. Points $A_2, C_2$ are the reflections of $B_1$ in lines $BC, AB$ respectively. Prove that lines $A_1A_2$ and $C_1C_2$ meet on the median of triangle $ABC$. Note that $A_1A_2 \parallel AC$. Let $M$ be the midpoint of $AC$ and let $X = A_1A_2 \cap AM$, $X’ = C_1C_2 \cap AM$. Note that $\angle A_1XB = \angle CBX$. So, $BX = BA_1$. Similarly, $BX’ = BC_1$. But $BA_1 = BC_1$. So, $X \equiv X’$. $\blacksquare$.