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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Solution

YES $△ ADB ≅ △ ADC (By SSS)$
AB = AC , DB = DC AND AD= DA

$∠BAD=∠CAD (c.p.c.t) ∠BAD+∠CAD = 40 ° 2 ∠BAD = 40 ° ∠ BAD = \frac{40 °}{2} = 20 °$

$∠ABC+∠BCA+∠BAC = 180 ° (Angle sum property) Since ∆ ABC is an isosceles triangle, ∠ ABC=∠BCA ∠ABC+∠ABC + 40 ° = 180 ° 2 ∠ABC = 180 ° - 40 ° = 140 ° ∠ ABC = \frac{140 °}{2} = 70 °$

$∠DBC+∠BCD+∠BDC = 180 ° (Angle sum property) Since ∆ABC is an isosceles triangle, ∠DBC=∠BCD ∠DBC+∠DBC + 100 ° = 180 ° 2 ∠DBC = 180 ° - 100 ° = 80 ° ∠ DBC = \frac{80 °}{2} = 40 °$

$In ∆BAD, ∠ABD+∠BAD+∠ADB = 180 ° (Angle sum property) 30 ° + 20 ° + ∠ADB = 180 ° (∠ABD=∠ABC-∠DBC) ∠ ADB = 180 ° - 20 ° - 30 ° ∠ ADB = 130 °$

$∠ A D B = 130 °$

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