In the given figure, `O` is the centre of the given circle, and chords `AB` and `CD` intersect at a point `E` inside the circle.
Prove that
`/_ AOC +/_BOD = 2 /_ AEC`

Explanation

Join `CB`.
We know that the angle made by `hat (AC)` at the centre O is double the angle made by it at a point B on the remaining part of the circle.
`:. /_AOC =2 /_ ABC`. ....(i)
Similarly, the anglemade by `hat (BD)` at the centre O is double the angle made by it at a point C on the remaining part of the circle.
`:. /_ BOD = 2 /_ BCD` ...(ii)
`:. /_ AOC + /_ BOD = 2 (/_ ABE + /_ BCD) ` [adding (i) and (ii) ]
` = 2 ( /_ EBC+ /_ BCE )`
`= 2 ( 180^(@) - /_CEB)` [ `:'` sum of `/_s` of `Delta ECB` is `180^(@)` ]
`=2 [ 180^(@) - ( 180^(@) - /_ AEC ) ]`
`= 2 /_ AEC`.
Hence, `/_ AOC + /_ BOD = 2 /_ AEC`.