✯Statement: The angle substended by an arc at the Centre of Cirlce is twice the angle substended by it on the remaining part of the circle.

✯Given: ∠AOB is the angle substended by arc AB at the centre of circle 'O' and ∠APB is the angle substended by arc AB on the remaining part of the circle

✯Required To Prove: ∠AOB = 2∠APB

✯Construction: Join PO and extend it to C.

✯Proof: Here there are 3 cases

• Arc AB is minor arc
• Arc AB is semi circular arc
• Arc AB is major arc

In all the 3 cases, In ∆OAP

☞ OA = OP (Radius of the Circle)

☞ ∠OPA = ∠OAP => x (Angles Opposite To Equal Sides of a ∆le)

Similarly, In ∆BOP

☞ OB = OP (Radius of the Circle)

☞ ∠OPB = ∠OBP => y (Angles Opposite To Equal Sides of a ∆le)

But, ∠AOC & ∠BOC are exterior angle of ∆OAP & ∆OBP respectively

So, ∠AOC = ∠OAP + ∠OPA = x+x ➝ 2x

∠BOC = ∠OPB + ∠OBP = y+y ➝ 2y

Thus, ∠AOB = 2x+2y

∠AOB = 2(x+y)

◕➜ Hence Proved

Hope It Helps You ✌️