Question

Answer this
In a quadrilateral ABCD, the Bisector of angle
$\angle \: C$
and
$\angle \: D$
intersect at o

Prove that
$\angle \: COD \: = ( \angle \: A \: + \angle \: B)$
​

Answer 1

Answer:

In ABCD, by angle sum property

∠A+∠B+∠C+∠D=360  

o

 

∴∠A+∠B=360  

o

−(∠C+∠D)

2

1

​  

(∠A+∠B)=180  

o

−  

2

1

​  

(∠C+∠D) ...... (1)

Now, In △COD

2

1

​  

∠C+  

2

1

​  

∠D+∠COD=180  

o

 

∠COD=180  

o

−  

2

1

​  

(∠C+∠D) ......... (2)

From (1) & (2)

∠COD=  

2

1

​  

(∠A+∠B)

Hence proved.

Step-by-step explanation: