`A B C D` is a cyclic quadrilateral in which: `B C A D ,\ /_A D C=110^0\ a n d\ /_B A C=50^0dot` Find `/_D A C` `/_D B C=80^0\ a n d\ /_B A C=40^0dot` Find `/_B C D` `/_B C D=100^0a n d\ /_A B D=70^0` find `/_A D B`

Explanation

`ABCD` is a cyclic quadrilateral
`AD || BC,/_ADC=110^0`
`/_BAC=50^0`
`/_B+/_D=180^0`(Sum of opposite angles)
`/_B+110^0=180^0`
`/_B=180^0-110^0`
`/_B=70^0`
Now in `/_\ ABC , /_CAB+/_ABC+/_BCA=180^0`
`50^0+70^0+/_BCA=180^0`
`/_BAC=180^0-120^0=60^0`
But `/_DAC=/_BCA`( alternate angles)
`/_DAC=60^0`
ii) `arc DC` subtends `/_DBC \and\ /_DAC` in the same segment
`/_DBC=/_DAC=80^0`
`/_DBC=/_DAC+/_CAB=80^0+40^0=120^0`
`/_DAB+/_BCD=180^0`(Sum of opposite angles of a cyclic quadrilateral).
`120^0+/_BCD=180^0`
`/_BCD=180^0-120^0=60^0`
iii) `ABCD` is cyclic quadrilateral and `BD` is joined.
`/_BCD =100^0 \and\ /_ABD=70^0`
`/_A+/_C=180^0`(Sum of opposite angles of cyclic quadrilateral)
`/_A+100^0=180^0`
`/_A=180^0-100^=80^0`
Now in `/_\ ABD,/_A+/_ABD+/_ADB=180^0`
`80^0+70^0+/_ADB=180^0`
`/_ADB=180^0-150^0`
`/_ADB=30^0`