Question

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that ∠CBD + ∠CDB = `1/2` ∠BAD

Answer 1

In a circle, ABCD is a quadrilateral having center A.

To prove ∠CBD + ∠CDB = `1/2` ∠BAD

Construction: Join AC.

Proof: As we know that angle subtended by an arc at the center is double the angle subtended by it at point on the remaining part of the circle.

So, ∠CAD = 2∠CBD  ...(i)

And ∠BAC = 2∠CDB  ...(ii)

Now, adding equation (i) and (ii), we get

∠CAD + ∠BAC = 2(∠CBD + ∠CDB)

∠BAD = 2(∠CBD + ∠CDB)

Hence, ∠CBD + ∠CDB = `1/2` ∠BAD.