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.