Question

11. The adjoining figure shows a circle with centre 0, in which
arc AB = arc CD. Prove that angle APB = angle
CQD.​

Answer 1

Answer:

Consider triangle AOB and triangle OCD

OA=OB=OC=OD=radius

∠AOB=∠DOC=70

In △OCD

OD=OC

∠ODC=∠OCD

∠OCD+∠ODC+∠DOC=180

2∠OCD=180

o

−70

o

∠OCD=55

o

=∠ODC

Answer 2

Step-by-step explanation:

Consider triangle AOB and triangle OCD

OA=OB=OC=OD=radius

∠AOB=∠DOC=70

In △OCD

OD=OC

∠ODC=∠OCD

∠OCD+∠ODC+∠DOC=180

2∠OCD=180

o

−70

o

∠OCD=55

o

=∠ODC