Ninth Grade Mathematics: Angles Subtended by Arcs at Center vs Circle
The angle subtended by a circular arc is the angle formed between the rays connecting its endpoints to a reference point. The Central Angle Theorem is the master theorem of circle geometry, revealing that an arc subtends an angle at the center of the circle that is exactly twice the angle it subtends at any point on the remaining circumference.
Statement and Significance of the Central Angle Theorem
Let AB be an arc of a circle with center O. The arc subtends the central angle Angle AOB at the center, and subtends the inscribed angle Angle APB at any point P lying on the remaining part of the circle.
The Central Angle Theorem (Inscribed Angle Theorem):
The angle subtended by an arc at the center of a circle is double
the angle subtended by it at any point on the remaining part of the circle.
Angle AOB = 2 * Angle APB
P (Inscribed Angle)
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A Central B
Angle AOB
Angle AOB = 2 * Angle APB
This theorem holds universally across all three geometric cases: when the arc is a minor arc, when the arc is a semicircle, and when the arc is a major arc (in which case Angle AOB is a reflex angle).
Deductive Proof of the Theorem
To prove the Central Angle Theorem, we use an auxiliary line connecting point P through center O, applying the Exterior Angle Theorem to two isosceles triangles:
Given: Arc AB of circle C(O, r) subtending Angle AOB at center O,
and Angle APB at point P on the remaining circumference.
To Prove: Angle AOB = 2 * Angle APB.
Construction:
Draw line segment PO and produce it past center O to an exterior point Q.
Proof:
Step 1: Consider Triangle OAP:
- Side OA = Side OP = r (Radii of the same circle)
- Therefore, Triangle OAP is an isosceles triangle.
- By the Base Angles Theorem: Angle OPA = Angle OAP = x.
Step 2: Side PO of Triangle OAP is extended to point Q.
By the Exterior Angle Theorem:
Angle AOQ = Angle OPA + Angle OAP
Angle AOQ = x + x = 2 * x = 2 * Angle OPA (Equation 1)
Step 3: Consider Triangle OBP:
- Side OB = Side OP = r (Radii of the same circle)
- Therefore, Triangle OBP is an isosceles triangle.
- By the Base Angles Theorem: Angle OPB = Angle OBP = y.
Step 4: Side PO of Triangle OBP is extended to point Q.
By the Exterior Angle Theorem:
Angle BOQ = Angle OPB + Angle OBP
Angle BOQ = y + y = 2 * y = 2 * Angle OPB (Equation 2)
Step 5: Add Equation 1 and Equation 2:
Angle AOQ + Angle BOQ = 2 * Angle OPA + 2 * Angle OPB
Angle AOB = 2 * (Angle OPA + Angle OPB)
Angle AOB = 2 * Angle APB.
The proof is complete. The central angle is always twice the inscribed angle subtended by the same arc.
The Three Morphologies: Minor, Semicircle, and Reflex
The theorem applies consistently regardless of the length of the subtending arc:
Case 1 (Minor Arc AB):
Central angle Angle AOB is acute or obtuse.
Angle AOB = 2 * Angle APB.
Case 2 (Semicircular Arc AB):
Endpoints A, O, B form a straight diameter line, so central angle Angle AOB = 180 degrees.
180 = 2 * Angle APB ===> Angle APB = 90 degrees!
(This proves Thales' Theorem: the angle in a semicircle is a right angle).
Case 3 (Major Arc AB):
Central angle is reflex Angle AOB (between 180 and 360 degrees).
Reflex Angle AOB = 2 * Angle APB.
Chapter Exercises
Exercise 1: In a circle with center O, minor arc AB subtends an angle of 110 degrees at the center. A point P lies on the major arc. Calculate the measure of Angle APB.
Exercise 2: Points A, B, and C lie on a circle with center O such that Angle AOB = 90 degrees and Angle BOC = 120 degrees. If point P lies on the circle opposite arc AC, find the measure of Angle APC.
Exercise 3: Triangle ABC is inscribed in a circle with center O. If Angle BAC = 65 degrees, find the measure of Angle OBC.
Solutions and Explanations
Solution 1: We are given that minor arc AB subtends a central angle of Angle AOB = 110 degrees at the center O. Point P lies on the major arc (the remaining part of the circle). By the Central Angle Theorem, the angle subtended by an arc at the center is double the angle subtended at any point on the remaining circumference: Angle AOB = 2 * Angle APB. Substitute the known value: 110 = 2 * Angle APB. Divide by 2: Angle APB = 110 / 2 = 55 degrees.
Solution 2: The complete angle around the center O is 360 degrees. We are given Angle AOB = 90 degrees and Angle BOC = 120 degrees. The remaining central angle Angle AOC subtending minor arc AC is Angle AOC = 360 - (90 + 120) = 360 - 210 = 150 degrees. Point P lies on the remaining circumference of the circle. By the Central Angle Theorem: Angle AOC = 2 * Angle APC, so 150 = 2 * Angle APC. Dividing by 2 yields Angle APC = 75 degrees.
Solution 3: In inscribed triangle ABC, Angle BAC = 65 degrees is an angle subtended by arc BC on the circumference. By the Central Angle Theorem, the angle subtended by arc BC at the center O is double the inscribed angle: Angle BOC = 2 * Angle BAC = 2 * (65) = 130 degrees. Now consider triangle OBC. The sides OB and OC are radii of the circle (OB = OC = r), making triangle OBC an isosceles triangle with base angles Angle OBC = Angle OCB. By the Angle Sum Property in triangle OBC: Angle BOC + Angle OBC + Angle OCB = 180, so 130 + 2 * Angle OBC = 180. Subtract 130: 2 * Angle OBC = 50. Divide by 2: Angle OBC = 25 degrees.