15. ABCD is a cyclic quadrilateral in which AB = BC and ABC = 70°.
AD produced meets BC produced at the
point P, where APB = 30°.
Calculate
a) ADB
b) ABD

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°