Question

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

Answer 1

Answer:

∠COD=1/2(∠A+∠B)

Step-by-step explanation:

the bisector of ∠C and ∠D intersect at O

=> in ΔCOD

∠C/2  + ∠D/2  + ∠COD = 180°

=> (1/2)(∠C + D) = 180°

=> (∠C + ∠D )  + 2∠COD = 360°

=> 2∠COD = 360° - (∠C + ∠D )

in a quadrilateral ∠A + ∠B + ∠C + ∠D = 360°

=> 360° - (∠C + ∠D )  = ∠A + ∠B

=>  2∠COD =  ∠A + ∠B

=> ∠COD=1/2(∠A+∠B)

QED

Proved