Magic incenter property

[ins-]

In the triangle $ABC$ - $I$ is its incenter. $A'$, $B'$, $C'$ are the intersection points of the lines $AI$, $BI$, $CI$ with the corresponding triangle sides. If we denote with $r_{XYZ}$ the inradii of the triangle $XYZ$ prove that:
a) $r_{AIB'}+r_{BIC'}+r_{CIA'} = r_{AIC'}+r_{CIB'}+r_{BIA'}$
b) $r_{AIB'}^2+r_{BIC'}^2+r_{CIA'}^2 = r_{AIC'}^2+r_{CIB'}^2+r_{BIA'}^2$
c) $r_{AIB'}^n+r_{BIC'}^n+r_{CIA'}^n = r_{AIC'}^n+r_{CIB'}^n+r_{BIA'}^n$,
where $n$ is a natural number.

[ins-]

I'm sorry - the problem is wrong. It just seems to be correct. The sums are very close as values.