In circle o, diameter ADB and chord AC are drawn. Find the measure of arc BC. Please help

Arc BC is 64°

Step-by-step explanation:

The parameters given are;

∠CAB = 32°

We note that the measure of arc BC = ∠CDB

∠DCA = ∠CAB = 32° (Base angles of an isosceles triangle)

∠ACB = 90° (Angle subtended at the center = Twice angle subtended at the circumference)

∠ACB = ∠DCB + ∠DCA

∴ ∠DCB = ∠ACB - ∠DCA = 90° - 32° = 58°

∠DBC = ∠DCB = 58° (Base angles of an isosceles triangle)

∴ ∠CDB + ∠DBC + ∠DCB = 180° (Sum of interior angles of a triangle)

∠CDB  = 180° - (∠DBC + ∠DCB) = 180° - (58° + 58°) = 64°

∠CDB  = Arc BC = 64°

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