GA Q9MCQ2 marks, −0.67In the figure below, ∠DEC+∠BFC\angle DEC + \angle BFC∠DEC+∠BFC is equal to _____ (A)∠BCD−∠BAD\angle BCD - \angle BAD∠BCD−∠BAD(B)∠BAD+∠BCF\angle BAD + \angle BCF∠BAD+∠BCF(C)∠BAD+∠BCD\angle BAD + \angle BCD∠BAD+∠BCD(D)∠CBA+∠ADC\angle CBA + \angle ADC∠CBA+∠ADCSign in to save attempts.Show answer← PreviousNext →SolutionsDiscussionLoading…Sign in with Google
