Let $E$ be a tangent point of the incircle of the $\triangle ADC$ with a side $\overline{AD}$. Call $I$ and $J$ incenters of the $\triangle ABC$ and $\triangle ADC$ with radius $R$ and $r$, respectively. Obviously, $J$ lies on the segment $\overline{CI}$ and so $R>r$. Notice that $IJ=R+r > 2r = 2JE \implies \cos \angle IJE = \frac{JE}{JI} < \frac{1}{2}$, which means $\angle IJE > 60^\circ$ as $\angle IJE < 90^\circ$. Finally, $\angle ABC = 2\cdot (\angle AIC - 90^\circ) = 2\cdot (\angle IJE + \angle JEI - 90^\circ) = 2\angle IJE > 120^\circ$, as desired.