Question

ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that ∠CBD + ∠CDB = ½ ∠BAD.

Answer 1

Consider a point E on the circle and join BE, DE and BD

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference

So we get

∠BAD = 2 ∠BED

It can be written as

∠BED = ½ ∠BAD …… (1)

Consider EBCD as a cyclic quadrilateral

So we get

∠BED + ∠BCD = 180^o

We can write it as

∠BCD = 180^o – ∠BED

Substituting equation (1)

∠BCD = 180^o – ½ ∠BAD …… (2)

Consider △ BCD

Using the angle sum property

∠CBD + ∠CDB + ∠BCD = 180^o

By using the equation (2)

∠CBD + ∠CDB + 180^o – ½ ∠BAD = 180^o

So we get

∠CDB + ∠CDB – ½ ∠BAD = 180^o – 180^o

On further calculation

∠CDB + ∠CDB = ½ ∠BAD

Therefore, it is proved that ∠CDB + ∠CDB = ½ ∠BAD.