TDSM 8.5

$$AC = CA \Rightarrow ACB = CAB$$

$$BC = CB \Rightarrow ABC = ACB$$

$$\Rightarrow CAB = ABC$$ (1)

$$AC = CA \Rightarrow BAC = BCA$$

$$BC = CB \Rightarrow BCA = CBA$$

$$\Rightarrow CBA = BAC$$ (2)

(1) and (2) $$\Rightarrow CAB+CBA = ABC+BAC \Leftrightarrow C(AB+BA) = (AB+BA)C$$ (proved)