Proof of the Inscribed Angle Theorem

In geometry, an inscribed angle is the angle formed in the interior of a circle when two chords intersect on the circle. A central angle is an angle whose vertex is the center of the circle and whose arms intersect the circumference of the circle.

The inscribed angle theorem tells us if an inscribed angle and a central angle of a circle intercept the same arc, the inscribed angle is the half of the central angle.

To prove the theorem, consider 3 cases when the vertex is on one arm of the inscribed angle, the vertex is between the arms of the inscribed angle and the vertex is outside the inscribed angle.

Vertex is on one arm of the inscribed angle

Because Point O is the center of the circle, △OCB is an isosceles triangle.

∠BOC + 2\alpha = 180\degree
(1)

Because ∠BOC and ∠\beta are supplementary angles

∠BOC + \beta = 180\degree
(2)

Subtract equation (1) from equation (2), we get,

\beta = 2\alpha

Vertex is between the inscribed angle

In this case, construct the line by joining point C and point O, which intersects the circumference of the circle at point D.

For the isosceles triangle △BCO , we have

\beta_1 = 2\alpha_1
(1)

For the isosceles triangle △ACO , we have

\beta_2 = 2\alpha_2
(2)

Add equation (1) and (2), we get,

\beta_1+ \beta_2 = 2(\alpha_1+\alpha_2)

\because \beta = \beta_1+ \beta_2 and \alpha = \alpha_1+\alpha_2

\therefore \beta =2 \alpha

Vertex is outside the inscribed angle

Similar to above case, construct the line by joining point C and point O, which intersects the circumference of the circle at point D.

For the isosceles triangle △BCO , we have

\beta+ \beta_2 = 2(\alpha+\alpha_2)
(1)

For the isosceles triangle △ACO , we have

\beta_2 = 2\alpha_2
(2)

Subtract equation (2) from equation (1) , we get

\beta = 2\alpha

In summary, there are three cases when an inscribed angle and a central angle intersect the same arc, which are,

1. the vertex of the central angle is on the arm of an inscribed angle

2. or between the inscribed angle

3. or outside inscribed angle

In each of the cases, it's proved that the central angle is the double of the inscribed angle.

Steven Zheng posted 7 months ago

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