Question

in a quadrilateral ABCD, the bisector of <C <D interset at 0
prove thout <cod=$(<A+CB)​

Answer 1

Step-by-step explanation:

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

c

irc]

⇒ ∠COD=180

∘

−(∠1+∠2)

⇒ ∠COD=180

∘

−(

2

1

∠C+

2

1

∠D)

⇒ ∠COD=180

∘

−

2

1

(∠C+∠D)

⇒ ∠COD=180

∘

−

2

1

[360

∘

−(∠A+∠B)]

⇒ ∠COD=180

∘

−180

∘

+

2

1

(∠A+∠B)

∴ ∠COD=

2

1

(∠A+∠B)