(i) Here,
∠BAF + ∠FAD = 180° (Linear pair)
∠FAD = 180° – ∠BAF = 180° – 90° = 90°
Also from the figure,
∠AFE = ∠ADF + ∠FAD (Exterior angle property)
∠ADF + 90° = 130°
∠ADF = 130° − 90° = 40°
(ii) We know that the sum of all the angles of a triangle is 180°.
Therefore, for △BDE, we have
∠BDE + ∠BED + ∠DBE = 180°
∠DBE = 180° – ∠BDE
∠BED = 180° – 90° – 40° = 50° …. Equation (i)
Again from the figure we have,
∠FAD = ∠ABC + ∠ACB (Exterior angle property)
90° = 50° + ∠ACB
∠ACB = 90° – 50° = 40°
(iii) From equation we have
∠ABC = ∠DBE = 50°