In the given figure, `AB` is a diameter of the circle and `CD||AB`. If `/_ DAB = 25^(@)`, calculate
`(i) /_ ACD ` and `/_ CAD`

Explanation

Join `AC` and `BD`.
We know that an angle in a semicircle is `90^(@)`.
`:. /_ ADB = 90^(@)`
Also, the sum of the angles of a triangle is `180^(@)`
`:. `in `Delta ` ABD , we have `/_ BAD + /_ ADB + /_ ABD = 180^(@)`
`implies 25^(@) + 90^(@) + /_ ABC = 180^(@)`
`implies /_ ABD = 180^(@) - 115^(@) = 65^(@)`
Since `ABDC` is a cyclic quadrilateral, wehave
`/_ABD + /_ ACD = 180^(@)implies /_ACD = 180^(@) - /_ ABD`
`implies /_ ACD = 180^(@) -65^(@) = 115^(@)`
Also , `AB ||CD` and AD is a transversal.
`:. /_ ADC = /_ BAD = 25^(@)` [alt. int. `/_ s `].
Now , in `Delta ` `ACD`, we have
`/_ ACD+/_ CAD +/_ ADC = 180^(@)`
`implies 115^(@) + /_ CAD + /_ 25^(@) = 180^(@)`
` implies /_ CAD = 180^(@) - 140^(@) = 40^(@)`