BD is the diameter of the circle. Points A, C are placed on a circle on opposite sides

BD is the diameter of the circle. Points A, C are placed on a circle on opposite sides of BD so that BC = 1 / 2BD, AC = AD. Prove that DB is the bisector of the angle ADC

BC = BD / 2 = BO (BC is equal to the radius).

BO = OS;

BC = OC.

ΔBCO – equilateral. <COB = 60 °.

<COD = 180 ° – <COB = 180 ° – 60 ° = 120 °.

Equal chords AC and AD rest on equal arcs AC and AD, therefore <COA = <DOA = (360 ° – <COD) / 2 = (360 ° – 120 °) / 2 = 120 °. All sides of the triangle CAD as chords rest on equal arcs. This means that ΔCAB is equilateral, and among others <CDA = 60 °.

<CDB rests on the arc CB, which corresponds to the central angle COB = 60 °, therefore <CDB = <COB / 2 = 60 ° / 2 = 30 °.

Answer. <CDB is <CDA / 2, hence BD is a bisector.