In Fig. 16.198, <

Given that,

A is the centre of the circle, then

AB = AD

ABCD is a parallelogram, then

AD ‖ BC, AB ‖ CD

CDE is a straight line, then

AB ‖ CE

Let,

∠BEC = ∠ABE = x’ (Alternate angle)

We know that,

The angle substended by an arc of a circle at the centre double the angle are angle substended by it at any point on the remaining part of circle

∠BAD = 2 ∠BEC

∠BAD = 2x’

In a rhombus opposite angles are equal to each other

∠BAD = ∠BCD = 2x’

Now, we have to find

=

=

=

Hence,

∠BCD: ∠ABE is 2: 1