[asy][asy] import graph; size(8cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -3.28, xmax = 11.75, ymin = -0.17, ymax = 8.12; draw((0.56,7.04)--(-2,2)); draw((-2,2)--(8,2)); draw((8,2)--(0.56,7.04)); draw(circle((1.33,4.04), 2.04)); draw((-0.21,4.52)--(2.48,5.74), red); draw((2.48,5.74)--(0.67,3.07), red); draw((2.24,0.51)--(2.48,5.74), red); draw((0.56,7.04)--(2.24,0.51)); draw((-0.49,4.97)--(2.24,0.51)); draw((-0.21,4.52)--(8,2)); dot((0.56,7.04),dotstyle); label("$A$", (0.66,7.18), NE * labelscalefactor); dot((-2,2),dotstyle); label("$B$", (-2.42-0.2,1.99), NE * labelscalefactor); dot((8,2),dotstyle); label("$C$", (8.1,2.14), NE * labelscalefactor); dot((1.33,4.04),dotstyle); label("$I$", (1.64,4.11), NE * labelscalefactor); dot((-0.49,4.97),dotstyle); label("$C_1$", (-1.03,5.04), NE * labelscalefactor); dot((1.33,2),dotstyle); label("$A_1$", (1.04-0.2,1.54-0.2), NE * labelscalefactor); dot((2.48,5.74),dotstyle); label("$B_1$", (2.58,5.88), NE * labelscalefactor); dot((2.24,0.51),dotstyle); label("$X$", (2.41,0.65), NE * labelscalefactor); dot((-0.21,4.52),dotstyle); label("$Y$", (-0.11,4.66), NE * labelscalefactor); dot((0.67,3.07),dotstyle); label("$Z$", (0.9,2.93), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy][/asy] First, $\angle AIC = 90^\circ + \frac{\angle B}{2}$, so $\angle AIC = \angle AC_1A_1$. (and $\angle CIA = \angle CA_1C_1$) Because of this, $\angle BC_1A_1 = \angle AIY$, so $A$, $I$, $Y$, and $C_1$ are concyclic. Also, since $CI$ is the perpendicular bisector of $B_1A_1$, $\angle YB_1C = \angle YA_1C = \angle AIC$, so $\angle AIY = 180^\circ - \angle AIC = 180^\circ - \angle YA_1C = 180^\circ - \angle YB_1C = \angle AB_1Y$, so $A$, $B_1$, $I$, and $Y$ are concyclic, so $A$, $B_1$, $I$, $Y$, and $C_1$ are concyclic. Similarly, $C$, $B_1$, $I$, $X$, and $A_1$ are concyclic. Because $I$ is on the perpendicular bisector of $C_1A_1$, $\angle IC_1A_1 = \angle IA_1C_1$, so $\angle YB_1Z = \angle IC_1Y = \angle IC_1A_1 = \angle IA_1C_1$ $=\angle IA_1Y = 180^\circ-\angle IA_1X = \angle XB_1Z$.

Why complicated? $ Y $ is on the perpendicular bisector of $ \angle C $, $ \angle A_{1}YC = \angle B_{1}YC $. Similarly, $ X $ is on the perpendicular bisector of $ \angle A $, $ \angle B_{1}XA = \angle C_{1}XA $. Therefore, $ I $ is also the incenter of triangle $ B_{1}XY $, and $ \angle Y B_{1}Z =\angle XB_{1}Z $. Prove that the two triangles $ ABC $ and $ YB_{1}X $ are similar.

We have the well-known problem below: Given a triangle $ABC$. The incenter $(I)$ of this triangle touches $AB$ and $AC$ at $M, N$, respectively. Denote $P$ is the intersection of $BI$ and $MN$. Prove that $\angle BPC = 90^0$ Proof: We have $(IC,IP) \equiv (NA,NM) (mod \pi)$ That means: $(NC,NP) \equiv (IC,IP) (mod \pi)$. So $N, I, C, P$ are on a circle $\Rightarrow CP \perp BP$ Return to the problem: Using this problem we have $\angle{AYC} = \angle{AXC} = 90^0$ So 4 points $Y, X, A, C$ are on a circle. $\Rightarrow \angle{XYC} = \angle{XAC} = \angle{IYB_1}$ (because $I, Y, B_1, A$ are on a circle) That means $YI$ is the bisector of $\angle{XYB_1}$ Similarly, $XI$ is the bisector of $\angle{YXB_1}$ It follows $I$ is the incenter of triangle $B_1XY$, we obtain $B_1I$ is the bisector of $\angle{YB_1X}$,we done!

Nice Problem! Let $M,N$ be the foot of the perpendicular from $Y$ to $BC$, and $Y$ to $CA$ respectively. Then, since $Y$ lies on the angle bisector of $\angle C$, $YM = YN$, $IA_1 = IB_1$. Thus we conclude that $\angle YB_1A = \angle YA_1B = 90 - B/2$. (**) (Note that $\angle IA_1C_1 = B/2$ !!) $(*)$ Similarly, from $X$, let $P$ and $Q$ be the foot of the perpendiculars from $X$ to sides $AB$ and $AC$, respectively. In a similar fashion we can derive $\angle XC_1B = \angle XB_1C$. Thus $\angle XC_1B = \angle XB_1C = 90 - B/2$ $(1)$ (Note that $\angle BC_1A_1 = 90-B/2$) Therefore by $(*)$ and $(1)$, we get that points $I,B_1,X,A_1,C$ to be concyclic. Hence $\angle ZB_1X = \angle IB_1X = \angle IA_1Z = B/2 = \angle YB_1Z$ (from (**)) Hence result follows.

Proof. Let $A_1C_1$ cuts $BC$ at $B_2$ we have $-1=(ACB_1B_2)=I(ACB_1B_2)=(XYZB_2)=B_1(XYZB_2)$. Note that $B_1B_2\perp B_1Z $ so $B_1Z$ is bisector of $\angle YB_1X$. We are done. General problem. Let $ABC$ be a triangle $P$ is a point such that pedal triangle $DEF$ of $P$ is also cevian triangle. $PB,PC$ cut $EF$ at $X,Y$, repectively. Prove that $DP$ is angle bisector of $\angle XDY$.

Proof.