Question

In a quadrilateral ABCD,CO and DO are the bisector of <C and <D respectively.Prove that <COD=1/2 (<A+<B)​

Answer 1

Step-by-step explanation:

In △COD,

∠COD+∠1+∠2=180 o

∠COD=180o −(∠1+∠2)

∠COD=180o − 1/2 (∠C+∠D)

∠COD=180-1/2 [360 −(∠A+∠B)]

∠COD= 1/2 (∠A+∠B)

Hope it will help you

Answer 2

Answer:

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