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Chord AB and CD intersect each other at point E in the initerior of the circle. In order to prove `/_CEB = (1)/(2) [m (arc AD ) + m (arc CB) ] `

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`/_ABD = (1)/(2)` m ( arc square ) …....(1)
`/_ CDB = (1)/(2) ` m ( arc CB …(2) …..( Inscribed angle theorem )
`/_ CEB ` is an exterior angle of `Delta DEB `
`/_CEB = /_EBD + /_EDB ` …. Reason `:`
i.e. `/_CEB = /_ABD + /_CDB ` ...(A-E-B,C-E-D )
`:. /_CEB = (1)/(2)` m (arc CB ) `+ (1)/(2) m arc square ) ` ....[From (1) and (2)]
`:. /_CEB = (1)/(2) [ m (arc square ) + m ( arc AD ) ]`

Activity `:`
`/_ABD = (1)/(2)` m ( arc AD ) ....(1)
`/_CDB = (1)/(2) m (arc CB ) ` ....(2)) .....( Inscribed angle theorem )
`/_CEB` is an exterior angle of `Delta DEB `
`/_CEB = /_EBD + /_EDB ` ....Reason `:` Remote interior angles
i.e. `/_CEB = /_ABD + /_ CDB ` (A-E-,C-E-D )`
` :. /_CEB = (1)/(2) m ( arc CB ) + (1)/(2) m (arc AD ) ` .....[From (1) and (2) ]
`:. /_CEB = (1)/(2) ` [ m ( arc (CB ) + m (arc AD ) ].
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