Question

In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = (∠A+∠B).

Answer 1

Given, 

In quadrilateral ABCD, 

CO is the bisector of ∠C

DO is the bisector of ∠D 

In ΔCOD,

\(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠D + ∠COD = 180°

⇒ ∠COD = 180° - \(\frac{1}{2}\)(∠D+∠C)

⇒ ∠D+∠C = 360 – (∠A+∠B)

So,

⇒ ∠COD = 180° - \(\frac{1}{2}\)(360 -(∠A+∠B))

⇒ ∠COD = \(\frac{1}{2}\) (∠A+∠B) Proved.