#### Question

In the following figure, AB is the diameter of a circle with centre O.

If chord AC = chord AD, Prove that:

(i) arc BC = arc DB

(ii) AB is bisector of ∠CAD.

Further, if the length of arc AC is twice the length of arc BC, find : (a) ∠BAC (b) ∠ABC

#### Solution

Given – In a circle with centre O, AB is the diameter and AC and AD are two chords such that AC = AD

To prove: (i) arc BC = arc DB

(ii) AB is the bisector of ∠CAD

(iii) If arc AC = 2 arc BC, then find

(a) ∠BAC

(b) ∠ABC

Construction: Join BC and BD

Proof : In right angled ∆ABC and ∆ABD Side AC = AD [Given]

Hyp. AB = AB [ Common]

∴ By right Angle – Hypotenuse – Side criterion of congruence

ΔABC ≅ ΔABD

(i) The corresponding parts of the congruent triangle are congruent.

∴ BC = BD [c.p.c.t]

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

(ii) ∠BAC = ∠BAD

∴ AB is the bisector of ∠CAD

(iii) If Arc AC = 2 arc BC,

Then ∠ABC = 2∠BAC

But ∠ABC + ∠BAC = 90°

⇒2∠BAC + ∠BAC = 90°

⇒ 3∠BAC = 90°

⇒` ∠BAC = (90°)/3= 30°`

∠ABC = 2∠BAC ⇒ ∠ABC = 2 ×30° = 60°