We will prove the inscribed angle theorem (inscribed angle = half the central angle).
📘 Example ① (Proof)
In a circle with center O and points A, B, and C on the circumference, show that ∠APB (P is the point on the circumference not on arc AB) is half of central angle ∠AOB.
Answer (Outline of Proof): Since OA=OB=OP (radii), △OAP and △OBP are isosceles triangles.
Property of exterior angles: ∠AOC = 2∠OAP = 2∠APC (property of isosceles triangles) (C is the intersection of the extension of OP and the circle).
Similarly, ∠BOC = 2∠BPC.
∠AOB = ∠AOC - ∠BOC = 2∠APC - 2∠BPC = 2∠APB.
In a circle with center O and points A, B, and C on the circumference, show that ∠APB (P is the point on the circumference not on arc AB) is half of central angle ∠AOB.
Answer (Outline of Proof): Since OA=OB=OP (radii), △OAP and △OBP are isosceles triangles.
Property of exterior angles: ∠AOC = 2∠OAP = 2∠APC (property of isosceles triangles) (C is the intersection of the extension of OP and the circle).
Similarly, ∠BOC = 2∠BPC.
∠AOB = ∠AOC - ∠BOC = 2∠APC - 2∠BPC = 2∠APB.
💡 Key Point
- Tools used in the proof: properties of isosceles triangles and the exterior angle theorem
- The key is to draw auxiliary lines (radii) from the center.
Practice Problems
- For semicircular arc AB, how many degrees is inscribed angle ∠APB at point P on the circumference? Explain using the central angle.
Answer and Explanation
- Answer: 90°. Central angle of a semicircle = 180°. Inscribed angle = 180 ÷ 2 = 90°.