TDSM 8.5

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[math]AC = CA \Rightarrow ACB = CAB[/math]

[math]BC = CB \Rightarrow ABC = ACB[/math]

[math]\Rightarrow CBA = ABC[/math] (1)

[math]AC = CA \Rightarrow BAC = BCA[/math]

[math]BC = CB \Rightarrow BCA = CBA[/math]

[math]\Rightarrow CBA = BAC[/math] (2)

(1) and (2) [math]\Rightarrow CAB+CBA = ABC+BAC \Leftrightarrow C(AB+BA) = (AB+BA)C[/math] (proved)