Let $I$ be the incenter and $A_2, B_2$ be the incircle tangency points on $CB, CA$. Let $M, N$ be the midpoints of $CB, CA$. Define $X=(ABC) \cap (CI)$ and let $X_1$ be the midpoint of $CX$. Note that $X, I, C'$ collinear. Claim: $\triangle XA_2B_2 \sim \triangle CA_1B_1$ directly similar. Proof: As $X$ is spiral center $BA \mapsto A_2B_2$ we have $\triangle XA_2B_2 \sim XBA \sim X_1MN$ are directly similar. Now recall that $M$ is the midpoint of $A_1A_2$ and $N$ is the midpoint of $B_1B_2$ and $X_1$ is the midpoint of $CX$, so by a spiral lemma we have $XA_2B_2 \sim \triangle CA_1B_1$. $\Box$ Claim: $CX$ is tangent to $(CA_1B_1)$. Proof: Observe that $\measuredangle XCB_1 = \measuredangle XA_2B_2 = \measuredangle CA_1B_1$. $\Box$ Let $\ell$ be the reflection of $CX$ over $CI$. Claim: $\ell \parallel A_1B_1$ Proof: Angle chase with similar triangles again. Or, take a $\sqrt{ab}$ inversion followed by a reflection about $CI$, and let $(\cdot)'$ denote the image of $(\cdot)$. Then $\ell$ is parallel to $A_1'B_1'$ by the second claim. But $A_1'B_1' \parallel A_1B_1$, as needed. $\Box$ Finally, let $K \in (ABC)$ be the midpoint of $II_C$ and let $C''$ be the reflection of $C'$ over $K$. Observe that $I_CC'' \parallel C'I \perp CX$ since $IC'I_CC''$ is a rhombus. Reflecting over $CI$ gives $I_CC' \perp \ell \parallel A_1B_1$, so we are done!

Weird Russian \(C-\) naming. Let \(CI \cap \odot ABC = M, C'I \cap \odot ABC \cap \odot (I) = D\), the incircle \(\odot (I) \cap BC, AB = E,F\) resp. and let \(B_e\) denote the bevan point. We know that \(CB'A' \sim \Delta DFE \Rightarrow \angle DIC = \angle B_eCI_c\) but since, \(C'M \perp\) bisects \( II_c\) \(\Rightarrow \angle DIC = \angle CI_CC'\) and hence the conclusion \(\quad \blacksquare\).