In a quadrilateral `A B C D ,A O` and`B O` are the bisectors of `/_A` and `/_B` respectively. Prove that `/_A O B=1/2(/_C+/_D)dot`

Explanation

Given=> `AO` and `BO` are the bisectors of∠
A
and ∠
B

`therefore ∠A =∠1+ ∠4 and ∠1=∠4->"(1)"`

`∠B=∠3 + ∠5 and ∠3=∠5`....(1)
To prove:∠`AOB=1/2(∠C+∠D)`
=>Proof:`∠A+∠B+∠C+∠D=360^0` - (angle sum property)
`therefore 1/2∠A+1/2∠B+1/2∠C+1/2∠D=360/2`
`1/2∠A+1/2∠B+1/2∠C+1/2∠D=180^0`....(eq-2)
In triangle AOB

`∠1+∠2+∠3=180^0`....(eq 3)
Equating (eq 2) and (eq 3)
`1/2∠A+1/2∠B+1/2∠C+1/2∠D=∠1+∠2+∠3`
`∠1+∠3+1/2∠C+1/2∠D=∠1+∠2+∠3`
`1/2∠C+1/2∠D=∠2`

`∠2=1/2(∠C+∠D)`

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