Question:
In a quadrilateral ABCD, CO and Do are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = 1/2(∠A and ∠B).
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 LC + 1/2 LD] [∵ OC and OD are bisectors of LC and LD respectively]
⟹ ∠COD = 180 – 1/2(LC + LD) ... (i)
In quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360° [Angle sum property of quadrilateral]
∠C + ∠D = 360° − (∠A + ∠B) .... (ii)
Substituting (ii) in (i)
⟹ ∠COD = 180 – 1/2(360 − (∠A + ∠B))
⟹ ∠COD = 180 − 180 +1/2(∠A + ∠B))
⟹ ∠COD = 1/2(∠A + ∠B))