SSC CHSL 15 Nov 2015 Shift 1 Paper

Given : $\angle \mathit{A}\mathit{B}\mathit{C}=75°\mathrm{and}\mathit{A}\mathit{D}\parallel \mathit{B}\mathit{C}$
To find : $\angle \mathit{B}\mathit{C}\mathit{D}=?$
Solution : In a cyclic quadrilateral, sum of opposite angles is supplementary.
⇒ $\angle \mathit{A}\mathit{B}\mathit{C}+\angle \mathit{A}\mathit{D}\mathit{C}=180°$
⇒ ∠ADC = 180°− 75°= 105°
Also, ∠ADC + ∠BCD = 180° [Angles on the same side of transversal]
⇒ ∠BCD = 180° − 105° = 75°