Answer:
a) ∠ADB is 55°
b) ∠ABD is 45°
Step-by-step explanation:
a) In the cyclic quadrilateral ABCD, we have;
Segment AB = Segment BC
∠ABC = 70°
Therefore, ∠ADC = 180° - 70° = 110° (Opposite angles are supplementary)
∠ADC + ∠CDP = 180° (Sum of angles on a straight line)
∴ ∠CDP = 180° - ∠ADC
∠CDP = 180° - 110° = 70°
∠DCP = 180° - 70° - 30° = 80°, (Angle sum property)
Similar to ∠DCP = ∠DAB = 80° (Exterior angle of a cyclic quadrilateral)
∠CAB = ∠ACB = (180° - 70°)/2 = 55° (Base angles of isosceles triangle ΔABC)
∠ADB = ∠ACB = 55° (Inscribed angle of a circle subtended by the same chord)
∠ADB = 55°
b) ∠ABD = 180° - ∠DAB - ∠ADB
∴ ∠ABD = 180° - 55° - 80° = 45°
∠ABD = 45°
