In the given figure, AB is the diameter of a circle with center O. If chord AC = chord AD, prove that :

(i) arc BC = arc DB

(ii) AB is the bisector of ∠CAD.

Further, if the length of arc AC is twice the length of arc BC, find :

(a) ∠BAC

(b) ∠ABC

Join BC and BD.

In △ABC and △ABD,

⇒ AC = AD [Given]

⇒ AB = AB [Common]

⇒ ∠ACB = ∠ADB [Both = 90°, Angle in semi-circle is a right angle.]

∴ △ ABC ≅ △ ABD [By RHS axiom of congruency]

(i) Since,

△ABC ≅ △ABD

∴ BC = BD [By C.P.C.T.]

∴ Arc BC = Arc BD [Equal chords have equal arcs].

Hence, proved that arc BC = arc BD.

(ii) Since,

△ABC ≅ △ABD

∴ ∠BAC = ∠BAD [By C.P.C.T.]

∴ AB is the bisector of ∠CAD.

Hence, proved that AB is the bisector of ∠CAD.

(a) Given,

arc AC = 2 arc BC

∴ ∠ABC = 2∠BAC

But,

⇒ ∠ABC + ∠BAC = 90° [ΔABC is right angled at C]

⇒ 2∠BAC + ∠BAC = 90°

⇒ 3∠BAC = 90°

⇒ ∠BAC = $\dfrac{90°}{3}$ = 30°.

Hence, ∠BAC = 30°.

(b) ∠ABC = 2∠BAC = 2 x 30° = 60°.

Hence, ∠ABC = 60°.