Relationship Between Central Angle and Inscribed Angle

Central angle = Angle subtended by an arc of the circle from the center of the circle.
Inscribed angle = Angle subtended by an arc of the circle from any point on the circumference of the circle. Also called circumferential angle and peripheral angle.

Figure below shows a central angle and inscribed angle intercepting the same arc AB. The relationship between the two is given by

$\alpha = 2\theta \, \text{ or } \, \theta = \frac{1}{2}\alpha$

if and only if both angles intercepted the same arc. In the figure below, θ and α intercepted the same arc AB.

Click here for the proof of the relationship.

Some Applications of the Relationship
• Right Triangle Inscribed in a Circle

The hypotenuse of triangle inscribed in a circle coincides with the diameter of the circle.

We can also say that an angle inscribed in a semicircle is a right angle. From the figure above, the diameter AC is the hypotenuse of triangles AB₁C, AB₂C, AB₃C, and AB₄C.

• Intersecting Chords

From the figure below, chords AC and BD intersect at E. Angle DAC and angle DBC intercepted the same arc CD, therefore, both angles are equal to one-half of the central angle DOC (not shown in the figure). We denote θ for angles DAC and DBC. Angle β = angle ACB = angle ADB, intercepting the arc AB. Triangle ADE is therefore similar to triangle BCE. By ratio and proportion of these similar triangles

$\dfrac{\text{opposite to } \theta}{\text{opposite to } \beta} = \dfrac{DE}{AE} = \dfrac{CE}{BE}$

$BE \times DE = AE \times CE$

This means that for intersecting chords in a circle, the product of segments of one is equal to the product of segments of the other.

• Intersecting Secants

Secant lines ED and EC intersect at point E as shown below. Angles ADB and ACB intercepted the same arc AB, therefore the angles are equal and we denote both by β. Also, angles DAC and DBC intercepted a common arc CD, both angles are equal and denoted as θ. Finally, angles EAC and EBD are both 180° - θ and denoted as Ø.

Therefore, triangles EAC and EBD are similar, and by ratio and proportion of similar triangles

$\dfrac{\text{opposite to } \varphi}{\text{opposite to } \beta} = \dfrac{DE}{BE} = \dfrac{CE}{AE}$

$DE \times AE = CE \times BE$

Also note that
$\varphi = \frac{1}{2}(\theta - \beta)$

• Intersecting Tangent and Secant

Tangent EB intersect to secant EC at point E as shown below. Angle BCE is equal to angle ABE, both are denoted by β.

Triangle ABE is similar to triangle BCE. By ratio and proportion

$\dfrac{\text{opposite to } \varphi}{\text{opposite to } \beta} = \dfrac{BE}{AE} = \dfrac{CE}{BE}$

$BE^2 = CE \times AE$

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