Inscribed Angles
General Rules & Definitions
1. Inscribed Angle Definition
An inscribed angle is an angle formed by two chords in a circle which have a common endpoint on the circumference.
A: When two chords intersect/meet at a point on the circle’s circumference, the angle so formed is an inscribed angle.
2. Central Angle Definition
A central angle is an angle whose vertex is at the center of the circle and whose legs (radii) intersect the circle at two distinct points.
A: When two radii meet at the center point $O$ of the circle, the angle formed at the center is a central angle.
3. Angle & Arc Relationship Summary
- Inscribed Angle: Always equal to half the degree measure of its opposite intercepted arc ($\angle APB \stackrel{\circ}{=} \frac{1}{2} \widehat{AB}$).
- Central Angle: Always equal to the degree measure of its opposite intercepted arc ($\angle AOB \stackrel{\circ}{=} \widehat{AB}$).
Interactive Concept Explorer
Drag points A or B to adjust arc $\widehat{AB}$. Drag P along circumference. Click the button below to snap/reset to Semicircle (Central Angle $180^\circ$, Inscribed Angle $90^\circ$).
Inscribed Angle Theorem Statement
Prove that the inscribed angles standing on the same arc are equal.
Given:
$O$ is the center of the circle.
$\angle APB$ and $\angle AQB$ are inscribed angles standing on the same arc $\widehat{AB}$.
To Prove: $\angle APB = \angle AQB$
Table
| Statements | Reasons |
|---|---|
| 1. $\angle APB \stackrel{\circ}{=} \frac{1}{2} \widehat{AB}$ | 1. An inscribed angle is half of its intercepted arc. |
| 2. $\angle AQB \stackrel{\circ}{=} \frac{1}{2} \widehat{AB}$ | 2. Same as reason 1. |
| 3. $\angle APB = \angle AQB$ | 3. From statements 1 and 2. |
Proved.
Figure
Inscribed Angle Theorem Statement
Prove that the inscribed angles standing on equal arcs of a circle are equal.
Given:
$O$ is the center of the circle.
Arc $\widehat{AB} = \text{Arc } \widehat{CD}$. Inscribed angles $\angle APB$ and $\angle CQD$ stand on arc $\widehat{AB}$ and arc $\widehat{CD}$ respectively.
To Prove: $\angle APB = \angle CQD$
Table
| Statements | Reasons |
|---|---|
| 1. $\angle APB \stackrel{\circ}{=} \frac{1}{2} \widehat{AB}$ | 1. An inscribed angle is half of its intercepted arc. |
| 2. $\angle CQD \stackrel{\circ}{=} \frac{1}{2} \widehat{CD}$ | 2. Same as reason 1. |
| 3. $\widehat{AB} = \widehat{CD}$ | 3. Given. |
| 4. $\angle APB = \angle CQD$ | 4. From statements 1, 2, and 3. |
Proved.
Figure
Question: Verify experimentally that the inscribed angles standing on the same arc are equal.
1. Figures:
Figure 1 ($r = 3\text{ cm}$)
Figure 2 (Intersecting Angles) ($r = 4\text{ cm}$)
Table
| Figure No. | $\angle APB$ | $\angle AQB$ | Result |
|---|---|---|---|
| 1 | $45^\circ$ | $45^\circ$ | $\angle APB = \angle AQB$ |
| 2 | $58^\circ$ | $58^\circ$ | $\angle APB = \angle AQB$ |
Question: Verify experimentally that the inscribed angles standing on equal arcs are equal.
1. Figures:
Figure 1 ($r = 3\text{ cm}$, $\widehat{AB} = \widehat{CD}$)
Figure 2 ($r = 4\text{ cm}$, $\widehat{AB} = \widehat{CD}$)
Table
| Figure No. | $\angle APB$ | $\angle CQD$ | Result |
|---|---|---|---|
| 1 | $38^\circ$ | $38^\circ$ | $\angle APB = \angle CQD$ |
| 2 | $42^\circ$ | $42^\circ$ | $\angle APB = \angle CQD$ |
Click and drag points A, B, P, or Q along the circle circumference.
Click and drag points A, B, C, P, or Q. Point D adjusts automatically so $\widehat{AB} = \widehat{CD}$.
Drag points A, B, or P. Click “Snap to 180°” for Semicircle demonstration!
Visual Proof: Diameter $PO$ divides the angle into two Isosceles Triangles ($\Delta AOP$ & $\Delta BOP$). The Exterior Angle Theorem proves $\angle AOB = 2 \times \angle APB$.
In $\Delta AOP$, $OA = OP = \text{radius} \implies \angle OAP = \angle OPA = x$.
Exterior angle $\angle AOC = x + x = 2x$.
Similarly, $\angle BOC = y + y = 2y$.
Therefore, Central Angle $\angle AOB = 2(x + y) = 2 \times \text{Inscribed Angle } \angle APB$.