Answer :
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
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