Question

pls someone answer this for free 50 points​

Answer 1

Consider the following figure, in which an arc (or segment) AB subtends ∠AOB at the center O, and ∠ACB at a point C on the circumference.

We have to prove that. Draw the line through O and C, and let it intersect the circle again at D, as shown:

We make the following observations:

• InΔOAC, ∠OAC=∠OCA              (because OA = OC )

• InΔOBC, ∠OBC=∠OCB              (because OB = OC )

Thus,

• ∠AOD = 2 × ∠ACO  
• ∠DOB = 2 × ∠OCB

⇒  ∠AOD + ∠DOB = 2(∠ACO + ∠OCB)

⇒  ∠AOB = 2 × ∠ACB

Hence Proved !!

Answer 2

Step-by-step explanation:

Theorem: The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference. The proof of this theorem is quite simple, and uses the exterior angle theorem – an exterior angle of a triangle is equal to the sum of the opposite interior angles.