Question

Fill In The Blanks

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

Answer 1

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


We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠CAD = 2∠CBD ...(1)
Also, ∠CAB = 2∠CDB ...(2)

Adding (1) and (2), we get
∠CAD + ∠CAB = 2∠CBD + 2∠CDB
⇒ ∠BAD = 2∠CBD + 2∠CDB
⇒ ∠BAD = 2(∠CBD + ∠CDB)
⇒ ∠CBD + ∠CDB = $\frac{1}{2}$∠BAD

Hence, k = $\underline{\frac{1}{2}}$.