Question

In Fig. 16.198, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD:∠ABE

Answer 1

Given : A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line.

To find : ∠BCD : ∠ABE

Solution :  

We  have A is the centre of the circle, then

AB = AD (Radius of a circle)

Since ABCD is a parallelogram, then  AD ‖ BC, AB ‖ CD

CDE is a straight line, then AB ‖ CE

Let, ∠BEC = ∠ABE = x  (Alternate angle)

Since, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∠BAD = 2 ∠BEC

∠BAD = 2x

Since, ABCD is a parallelogram then the opposite angles are equal to each other.

∠BAD = ∠BCD = 2x

Now, we have to find the ratio of ∠BCD : ∠ABE  :

∠BCD : ∠ABE = 2x : x

∠BCD : ∠ABE  = 2 : 1

Hence, ∠BCD : ∠ABE is 2 : 1

 

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Answer 2

Answer:

Step-by-step explanation:

AB = AD (Radius of a circle)

Since ABCD is a parallelogram, then  AD ‖ BC, AB ‖ CD

CDE is a straight line, then AB ‖ CE

Let, ∠BEC = ∠ABE = x  (Alternate angle)

Since, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∠BAD = 2 ∠BEC

∠BAD = 2x

Since, ABCD is a parallelogram then the opposite angles are equal to each other.

∠BAD = ∠BCD = 2x

Now, we have to find the ratio of ∠BCD : ∠ABE  :

∠BCD : ∠ABE = 2x : x

∠BCD : ∠ABE  = 2 : 1