Question

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

If chord AC = chord AD, prove that:

1. arc BC = arc DB
2. AB is bisector of ∠CAD.

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

1. ∠BAC
2. ∠ABC

Answer 1

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

To prove:

1. arc BC = arc DB
2. AB is the bisector of ∠CAD
3. If arc AC = 2 arc BC, then find

1. ∠BAC
2. ∠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^circ)/3 = 30^circ`

∠ABC = 2∠BAC

`=>` ∠ABC = 2 × 30° = 60°