Question

In the given circle, O is the centre. Chords AD = CD, If ∠ABC = 40°, find ∠ABD, ∠BOC and ∠COD.

Answer 1

Given:

• O is the center of the circle.
• ∠ABC = 40°
• AB is a diameter.
• AB = CD ⇒ triangle ADC is isosceles.
• Required: Find ∠ABD, ∠BOC and ∠COD.

Step 1: ∠ABD

We are given:

• ∠ABC = 40°
• AB is a diameter, so triangle ABC is a right triangle at point C.

In triangle ABC:

∠ABC = 90°   ...(Angle in a semicircle)

∠CAB = 180° – 90° – 40° = 50°

Now in triangle ADC:

AD = CD ⇒ triangle is isosceles ⇒ ∠DAC = ∠DCA

Since point D lies on arc AC, ∠DCA is a part of angle ∠ABC.

We can say:

∠ABD = ∠ABC – ∠DBC = 40° – ∠DCA

If we suppose ∠DCA = 20°, then:

∠ABD = 40° – 20°

= 20°

Step 2: ∠BOC

Arc BC is subtended by ∠BAC at the circumference.

The central angle ∠BOC subtending the same arc is:

∠BOC = 2 × ∠BAC

= 2 × 50°

= 100°

Step 3: ∠COD

From triangle DAC:

• AD = CD ⇒ triangle is isosceles
• We assumed ∠DAC = ∠DCA = 20°

The central angle subtending arc CD is:

∠COD = 2 × ∠DCA

= 2 × 20°

= 40°