Question

3. In a quadrilateral ABCD, the bisector of ZC and
ZD intersect at 0.
Prove that ZCOD
-(ZA+ZB)​

Answer 1

Answer:

Step-by-step explanation:Angle COD= 1800- 1/2angle C-1/2 angle D=1800-1/2(angle C+D)----------(1)

Angle (A+B+C+D)=3600

AngleC+D= 3600 - AngleA- angle B--------------(2)

Substituting (2) in (1)

Angle COD= 1800-1/2(3600- angle A- Angle B)

Angle COD= 1800-1800+1/2 Angle A + 1/2 Angle B= 1/2(Angle A+ Angle B)