**Solution**

Let us denote $R$ to be the radius of $C_O$ and $r$ to be the radius of $C_I$. Also let $O$ and $I$ to be the centers of the circle. Let also Suppose $AB$ is tangent to $C_I$ at $L$. And suppose $AI$ meets $C_O$ at $M$ and $MO$ meets $C_O$ at $N$.

Now suppose $a = \angle MAZ = \angle MAB = \angle MNC = \angle MCB$. Because $\angle ALI = \angle MCN = 90^o$ then by similarity of $\triangle ALI$ and $\triangle MNC$ we have: $$AI.MC = MN.LI = 2Rr$$

Now $\angle MIC = \angle IAC + \angle ACI = x + \angle ACI = \angle MCB + \angle ICB = \angle MCI$ so that we know $MC = MI$. Thus: $$AI.MI = 2Rr$$

Now, we draw similar figures to the triangle $X,Y,Z$. That is: $XY$ is tangent to $C_I$ at $L'$. And suppose $XI$ meets $C_O$ at $M'$ and $M'O$ meets $C_O$ at $N'$.

In our analysis above, we showed that $AI.MC = 2Rr$ but we didn't use the fact that $BC$ is tangent to $C_I$, so we can repeat the argument to claim that $$XI.M'Z = 2Rr$$.

On the other hand, $XI.IM' = AI.IM = 2Rr$ so we have $IM' = M'Z$ which means $IM'Z$ is an isosceles. $$\angle M'IZ = \angle M'ZI$$ $$\angle M'XZ + \angle IZX = \angle M'ZY + \angle YZI$$ Because $XY$ and $XZ$ are both tangent to $C_I$ then $XM$ is a bisector of $\angle YXZ$ so that $\angle M'XZ = \angle M'XY = \angle M'ZY$. So our equation above becomes: $$\angle IZX = \angle YZI$$ which means $ZI$ is also internal angle bisector, which means $YZ$ is tangent to $C_I$.

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