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ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If ∠BAC = 40° and ∠BDC = 100°, then find ∠ADB.

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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°

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