Question

In the circle, O is the centre, chords AE = AD = DC and ∠ABC = 56°. Find ∠CBD, ∠CBE, ∠BOC and ∠COD.

Answer 1

Given:

O is the centre of the circle.

AE = AD = DC (i.e. all three chords are equal)

∠ABC = 56°

We need find:

• ∠CBD
• ∠CBE
• ∠BOC
• ∠COD

1. Find ∠CBD

We are given:

∠ABC = 56°

Now point D lies between A and C and since chords AD = DC, triangle ΔADC is isosceles.

Therefore, point D is the midpoint of arc AC and ∠CBD is half of ∠ABC:

∠CBD = `56^circ/2`

= 28°

2. Find ∠CBE

This angle is made of two parts:

∠CBE = ∠CBD + ∠DBE

From above:

• ∠CBD = 28°
• ∠DBE = ∠ABC = 56°

So:

∠CBE = 28° + 56° = 84°

3. Find ∠BOC

Central angle ∠BOC subtends the same arc as ∠BAC, which is double the corresponding angle at the circumference:

∠BOC = 2 × ∠CBD

= 2 × 34°

= 68°

4. Find ∠COD

Similarly, angle at center ∠COD subtends arc CD and ∠COD = 2 × ∠CAD = 2 × 28° = 56°