In the given figure, AB is a diameter of a circle with centre O and `DO || CB`.
In `/_ BCD = 120^(@)`, calculate
`(i) /_ BAD` , (ii) `/_ ABD`
(iii) `/_ CBD`, (iv) `/_ ADC`
Also, show that `Delta AOD` is an equilateral triangle.

Answer

`(i) /_ BAD = 60^(@), (ii) /_ ABD = 30^(@) , (iii) /_ CBD = 30^(@) , (iv ) /_ ADC = 120^(@)`

Explanation

`/_ BCD + /_ BAD = 180^(@) implies BAD = 60^(@)`
`/_BAD = 90^(@)` [ angle in a semicircle ]
`/_ BDA + /_ BAD + /_ ABD = 180^(@) implies /_ ABD = 30^(@)`
`OD = OA implies /_ ODA = /_ OAD= /_ BAD = 60^(@)`
`:. /_ ODB = 90^(@) - /_ ODA = 30^(@)`
`:. /_ CBD= /_ ODB = 30^(@) implies /_ CDB = 180^(@) - ( 120^(@) + 30^(@) ) = 30^(@)`
and `/_ ADC= /_ ADB + /_ CDB= ( 90^(@) + 30^(@)) = 120^(@)`