In ΔBEC
∠BEC + ∠ECB +∠CBE = 180° [Sum of angles of a triangle is 180°]
90° + 40° + ∠CBE = 180°
∠CBE = 180°-130°
∠CBE = 50°
∠CBE = ∠ADC = 50° (Opposite angles of a parallelogram are equal)
∠B = ∠D = 50° [Opposite angles of a parallelogram are equal]
∠A + ∠B = 180° [Sum of adjacent angles of a triangle is 180°]
∠A + 50° = 180°
∠A = 180°-50°
So, ∠A = 130°
In ΔDFC
∠DFC + ∠FCD +∠CDF = 180° [Sum of angles of a triangle is 180°]
90° + ∠FCD + 50° = 180°
∠FCD = 180°-140°
∠FCD = 40°
∠A = ∠C = 130° [Opposite angles of a parallelogram are equal]
∠C = ∠FCE +∠BCE + ∠FCD
∠FCD + 40° + 40° = 130°
∠FCD = 130° – 80°
∠FCD = 50°
∴ ∠EBC = 50o, ∠ADC = 50o and ∠FCD = 50o
