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Question
In a quadrilateral ABCD, CO and DO are the bisectors of `∠`C and ∠D respectively. Prove that
`∠`COD = `1/2` (`∠`A+ `∠`B).
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Solution
In ΔDOC
`∠`1+ `∠`COD + `∠`2 =180° [Angle sum property of a triangle]
⇒ `∠`COD = 180 - `∠`1- `∠`2
⇒ `∠`COD =180 - `∠`1+ `∠`2
⇒ `∠`COD = 180- `[1/2∠c+1/2∠d]`
[ ∵ OC and OD are bisectors of `∠`C and `∠`D represents ]
⇒ `∠`COD = 180-`1/2` (`∠`C and `∠`D)] ............1
In quadrilateral ABCD
`∠`A + `∠`B + `∠`C + `∠`D = 360°
`∠`C + `∠`D = 360 - `∠`A + `∠`B ..............(2) [ Angle sum property of quadrilateral]
Substituting (ii) in (i)
⇒ `∠`COD = 180 -`1/2`(360 - `∠`A + `∠`B ))
⇒ `∠`COD = 180 -180 +`1/2`(`∠`A +`∠`B )
⇒ `∠`COD =`1/2`(`∠`A +`∠`B )
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