Answer :


In ΔABC,


A + B + C = 180° [Sum of all angles of a triangle = 180°]


According to the figure,


B + (α + DAC) + (γ + DCA) = 180°


DAC + DCA + α + β + γ = 180°


DAC + DCA = 180° - (α + β + γ) …. (i)


In ΔADC,


x + DAC + DCA = 180° [Sum of all angles of a triangle = 180°]


x = 180° - DAC - DCA


x = 180° - 180° + (α + β + γ)


x = (α + β + γ)


Hence proved.


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