In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = 1/2 (∠A+∠B).
Answers
Given : In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively.
To Prove : ∠COD = 1/2 (∠A+∠B).
Proof :
In quadrilateral ABCD ,
We know that , the sum of the angles of a quadrilateral is 360°.
∠A+∠B + ∠C + ∠D = 360°
∠C + ∠D = 360° - [∠A + ∠B] ..............(1)
In ΔCOD, we have :
We know that , the sum of the angles of a triangle is 180°.
∠OCD + ∠ODC + ∠COD = 180°
⇒ ½ ∠C + ½ ∠D + ∠COD = 180°
[CO and DO are the bisectors of ∠C and ∠D]
⇒ ½ [∠C + ∠D] + ∠COD = 180°
⇒ ∠COD = 180° - ½ [∠C + ∠D]
⇒ ∠COD = 180° - ½ [360° - [∠A + ∠B]
[From eq 1]
⇒ ∠COD = 180° - ½ × 360° + ½ [∠A + ∠B]
⇒ ∠COD = 180° - 180° + ½ [∠A + ∠B]
⇒ ∠COD = 180° - 180° + ½ [∠A + ∠B]
⇒ ∠COD = ½ [∠A + ∠B]
Hence Proved.
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Answer:-
Given :
In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively.
To Prove :
∠COD = 1/2 (∠A+∠B).
Proof :
In quadrilateral ABCD ,
We know that , the sum of the angles of a quadrilateral is 360°.
∠A+∠B + ∠C + ∠D = 360°
∠C + ∠D = 360° - [∠A + ∠B] ...(1)
In ΔCOD, we have :
We know that , the sum of the angles of a triangle is 180°.
∠OCD + ∠ODC + ∠COD = 180°
⇒ ½ ∠C + ½ ∠D + ∠COD = 180°
[CO and DO are the bisectors of ∠C and ∠D]
⇒ ½ [∠C + ∠D] + ∠COD = 180°
⇒ ∠COD = 180° - ½ [∠C + ∠D]
⇒ ∠COD = 180° - ½ [360° - [∠A + ∠B]
[From eq 1]
⇒ ∠COD = 180° - ½ × 360° + ½ [∠A + ∠B]
⇒ ∠COD = 180° - 180° + ½ [∠A + ∠B]
⇒ ∠COD = 180° - 180° + ½ [∠A + ∠B]
⇒ ∠COD = ½ [∠A + ∠B]
Hence Proved.
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