Answer:

It is given that AB is a DIAMETER of a circle with centre O and DO || CB

(i) We know that ABCD is a cyclic quadrilateral

It can be written as

∠BCD + ∠BAD = 180o

By substituting the values

120o + ∠BAD = 180o

On further calculation

∠BAD = 180o – 120o

By subtraction

∠BAD = 60o

(ii) We know that the angle in a semi-circle is RIGHT angle

∠BDA = 90o

Consider △ ABD

By using the angle sum property

∠BDA + ∠BAD + ∠ABD = 180o

By substituting the values

90o + 60o + ∠ABD = 180o

On further calculation

∠ABD = 180o – 90o – 60o

By subtraction

∠ABD = 180o – 150o

So we GET

∠ABD = 30O

(iii) We know that OD = OA

So we get ∠ODA = ∠OAD = ∠BAD = 60o

From the figure we know that

∠ODB + ∠ODA = 90o

By substituting the values

∠ODB + 60o = 90o

On further calculation

∠ODB = 90o – 60o

By subtraction

∠ODB = 30o

It is given that DO || CB

We know that the alternate angles are equal

∠CDB = ∠ODB = 30o

(iv) From the figure we know that

∠ADC = ∠ADB + ∠CDB

By substituting the values

∠ADC = 90o + 30o

By addition

∠ADC = 120o

Consider △ AOD

By using the angle sum property

∠ODA + ∠OAD + ∠AOD = 180o

By substituting the values

60o + 60o + ∠AOD = 180o

On further calculation

∠AOD = 180o – 60o – 60o

By subtraction

∠AOD = 180o – 120o

So we get

∠AOD = 60o

We know that all the angles of the △ AOD is 60o

Therefore, it is proved that △ AOD is an equilateral triangle.