Question

In the figure AB and CB are chords of a circle equidistant from the center O. Prove that diameter DB bisect angle ABC and angle ADC.

Answer 1

In ∆BAD and ∆BCD,
∠BAD = ∠BCD since 90° each,
as angle in a semicircle is a right angle,
BD = BD common
BA = BC Chords equidistant from the centre ar equal in length
⇒ ∆BAD ≅ ∆BCD RHS
⇒∠ABD = ∠CBD
∠ADB = ∠CDB CPCT
⇒diameter BD bisects ∠ABC and ∠ADC