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.

Inscribed and Central Angles


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.
Right triangle inscribed in a cirlce


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
Intersecting Chords


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

$BE \times DE = AE \times CE$

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.
Intersecting Secants


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

$\dfrac{opposite\,\,to\,\,\phi}{opposite\,\,to\,\,\beta} = \dfrac{DE}{BE} = \dfrac{CE}{AE}$

$DE \times AE = CE \times BE$

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.
Intersecting secant and tangent

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

$\dfrac{opposite\,\,to\,\,\phi}{opposite\,\,to\,\,\beta} = \dfrac{BE}{AE} = \dfrac{CE}{BE}$

$BE^2 = CE \times AE$