Question

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

  

Answer 1

It is given that ‘ABCD’ is a parallelogram. But since ‘A’ is the centre of the circle, the lengths of ‘AB’ and ‘AD’ will both be equal to the radius of the circle. 

So, we have  AB = AD .

Whenever a parallelogram has two adjacent sides equal then it is a rhombus.

So ‘ABCD’ is a rhombus.

Let `angleBDE = x° `.

We know that in a circle the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

By this property we have

`angleBAD = 2 (angleBDE )`

`angleBAD  = 2 x°` 

In a rhombus the opposite angles are always equal to each other.

So,  `angleBAD = angleBCE = 2x°`

Since the sum of all the internal angles in any triangle sums up to 180°  in triangle ΔBEC , we have

`angleEBC + angleABE = angleEBC` = 180°

                                   `angleEBC = 180° - angleBEC - angle BCE`

                                               = 180°  - x° -2x°

                                    `angleEBS` = 180° - 3x°  

In the rhombus ‘ABCD’ since one pair of opposite angles are ‘2x° ’ the other pair of opposite angles have to be  (180° - 2x°)

From the figure we see that,

`angleEBC + angle ABE = angleABC `

                  `angleABE = angleABC - angleEBC `

                                = 180° - 2x° - (180° - 3x°)

                    `angleABE` = x°

So now we can write the required ratio as,

`(angleBCD)/(angleABE) = (2x°)/(x°)`

`(angleBCD)/(angleABE) = 2/1`

Hence the ratio between the given two angles is 2: 1 .