Given ABC and DBC are both isosceles triangles on a common base BC
∠BAD = ∠CAD (corresponding parts of congruent triangles)
∠BAD + ∠CAD = 40°/ 2
∠BAD = 40°
∠BAD = 40°/2 =20°
∠ABC + ∠BCA + ∠BAC = 180° (Angle sum property)
Since ΔABC is an isosceles triangle,
∠ABC = ∠BCA
∠ABC +∠ABC + 40° = 180°
2 ∠ABC = 180°– 40° = 140°
∠ABC = 140°/2 = 70°
∠DBC + ∠ BCD + ∠ BDC = 180° (Angle sum property)
Since ΔABC is an isosceles triangle, ∠ DBC = ∠BCD
∠DBC + ∠DBC + 100° = 180°
2 ∠DBC = 180°– 100° = 80°
∠DBC = 80°/2 = 40°
In Δ BAD,
∠ABD + ∠BAD + ∠ADB = 180° (Angle sum property)
30° + 20° + ∠ADB = 180° (∠ADB = ∠ABC – ∠DBC),
∠ADB = 180°– 20°– 30°
∠ADB = 130°