Given : quadrilateral ABCD, in which ∠ABC + ∠ADC = 180° and ∠BAD + ∠BCD = 180°
To Prove : ABCD is a cyclic quadrilateral.
Construction : Draw a circle passing through points A, B, C but not D. But cuts AD at E. Join EC
Proof : quadrilateral ABCE is cyclic
Thus, ∠ABC + ∠AEC = 180° …..(i)
But given that
∠ABC + ∠ADC = 1800 ……(ii)
From equations (i) and (ii)
∠AEC = ∠ADC ……(iii)
But It is impossible because one is exterior and other is interior angle of ΔCED. it is possible only when D and E coincides so circle will pass through A, B, C and D
i.e., ABCD is a cyclic quadrilateral.