In a quadrilateral ABCD, O is a point inside the quadrilateral such that AO and BO are the bisectors of ∠A and ∠B respectively. Prove that ∠AOB =  
1
2
 (∠C + ∠D).


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Answer:

 

1
2
  (∠C + ∠D)

Step by Step Explanation:
  1. Following figure shows the quadrilateral ABCD,

    According to the question, AO and BO are the bisectors of ∠A and ∠B respectively.
    Therefore, ∠BAO =  
    ∠A
    2
     ,
    ⇒ ∠ABO =  
    ∠B
    2
     
  2. We know that the sum of all angles of a quadrilateral is equals to 360°.
    Therefore, ∠A + ∠B + ∠C + ∠D = 360°
    ⇒ ∠C + ∠D = 360° - (∠A + ∠B) -----(1)
  3. In ΔAOB,
    ∠BAO + ∠ABO + ∠AOB = 180°  [Since, we know that the sum of all three angles of a triangle is equals to 180°.]
    ⇒  
    ∠A
    2
      +  
    ∠B
    2
      + ∠AOB = 180°
    ⇒ ∠AOB = 180° -  
    ∠A
    2
      -  
    ∠B
    2
     
    ⇒ ∠AOB =  
    360° - ∠A - ∠B
    2
     
    ⇒ ∠AOB =  
    360° - (∠A + ∠B)
    2
     
    ⇒ ∠AOB =  
    (∠C + ∠D)
    2
       [From equation (1).]
  4. Hence, ∠AOB =  
    1
    2
     (∠C + ∠D)

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