In a ΔABC, ∠ B = 45°, ∠C = 55° and bisector of ∠ A meets BC at a point D. Find ∠ ADB and ∠ ADC.

Given: In Δ ABC, ∠ B = 45°, ∠ C = 55°. AD is the bisector of ∠ A meets BC at D
To find: ∠ ADB and ∠ ADC.
Proof: In Δ ABC, ∠ A + ∠ B + ∠ C = 180°
Þ ∠ A + 45° + 55° =180°
⇒ ∠ A = 180° - 100° = 80°
⇒ ∠ BAD = ∠ CAD = ∠ A [AD is the bisector of ∠ A]
                       = × 80° = 40°
In Δ ABD, ∠ ADB + ∠ DBA + ∠ BAD = 180° [Sum of the three angles of a triangle is 180° ]
⇒ ∠ADB + 45° + 40° =180°
        ⇒ ∠ADB + 85° = 180°
⇒ РADB = 180° - 85° = 95°
Now take angles РADB and ∠ ADC. They form a linear pair of angles.
          \\ 95° + ∠ ADC = 180°
⇒ ∠ ADC = 180° - 95° = 85°