Question

In a quadrilateral ABCD, the bisector of angle C and
D intersect at O
Prove that angle COD =1/2 (angle A+angle B)​

Answer 1

Answer:

In given figure ABCD is a quadrilateral.

⇒ CO and DO are bisector of ∠C and ∠D

In △COD,

⇒ ∠COD+∠1+∠2=180 ∘ (Sum of all interior angles of triangle is 180∘)

⇒ ∠COD=180∘−(∠1+∠2)

⇒ ∠COD=180∘ −∠2-∠1

⇒ ∠COD=180 ∘−( 1/2∠C+1/2∠D)

⇒ ∠COD=180 ∘− 1/2 [360 ∘ −(∠A+∠B)]

⇒ ∠COD=180 ∘−180 ∘+ 1/2(∠A+∠B)

∴ ∠COD= 1/2 (∠A+∠B)

Hence proved