In a quadrilateral `ABCD`, `CO` and `DO` are the bisectors of `/_C` and `/_D` respectively. Prove that `/_C O D=1/2(/_A+/_B)`

Explanation

In quadrilateral ABCD,
CO and DO are the bisectors of `angleC` and `angleD` respectively.
In `/_\COD`,
`angleDCO`+`angleCDO`+`angleCOD`=`180^@` (sum of angles of a triangle)
1/2`angleC`+1/2`angleD`+`angleCOD`=`180^@`
`angleCOD`=`180^@`-1/2(`angleC`+`angleD`)
=`180^@`-1/2(`angleC`+`angleD`)
But`angleA`+`angleB`+`angleC`+`angleD`=`360^@`=`angleC`+`angleD`=`360^@`-(`angleA`+`angleB`)
Hence,`angleCOD`=`180^@`-[`360^@`-`angleA`-`angleB`)
=`180^@`-`180^@`+1/2(`angleA`+`angleB`)
Hence Proved.