Question

in a quadrilateral, CO and DO are the bisectors of <c and <d respectively. prove that <A + <B = 2<COD​

Answer 1

Answer:

ANSWER

In △COD,

∠COD+∠1+∠2=180

o

∠COD=180

o

−(∠1+∠2)

∠COD=180

o

−

2

1

(∠C+∠D)

∠COD=180

o

−

2

1

[360

o

−(∠A+∠B)]

∠COD=

2

1

(∠A+∠B)