Inscribed angle theorem

It is necessary to prove that the inscribed angle is equal to half the arc on which it rests.

There are three options for the location of the inscribed angle:


An inscribed angle passes one side through the center of the circle

Inscribed angle theorem. An inscribed angle passes one side through the center of the circle
Inscribed angle theorem. An inscribed angle passes one side through the center of the circle. Proof
  • Let's connect points O and A. The radii of the circle are the same. This means that triangle ABO is isosceles, and angle 1 = angle 2. (An isosceles triangle has equal base angles.)
  • Angle AOC is the outer angle of triangle AOB. And the external angle is equal to the sum of two other angles not adjacent to it, angle AOC is equal to the sum of angle 1 and angle 2. Angle 1 and angle 2 are the same. Which means ∠ 1 = ½ ∠ AOC. And angle AOC is central and equal to the arc. This means that angle 1 (angle ABC) is equal to half of arc AC:

We have proven that an inscribed angle is equal to half the arc on which it rests.

The center of the circle is inside the inscribed angle

Inscribed angle theorem. The center of the circle is inside the inscribed angle
Inscribed angle theorem. The center of the circle is inside the inscribed angle. Proof
  • Let's draw ray BO and get two angles (angle 1 and angle 2), in which one side passes through the center of the circle. It was proven above that an inscribed angle is equal to half the arc on which it rests. So ∠ 1 = ½ AD. Angle 2 is also an inscribed angle; one side passes through the center of the circle. And it was also proven that this inscribed angle is equal to half the arc on which it rests. ∠ 2 = ½ DC.
  • Now, if you add angle 1 and angle 2, you get angle ABC. ∠ ABC = ∠ 1 + ∠ 2 = ½ AC.

We have again proven that an inscribed angle is equal to half the arc on which it subtends.

The center of the circle is outside the inscribed angle

Inscribed angle theorem. The center of the circle is outside the inscribed angle
Inscribed angle theorem. The center of the circle is outside the inscribed angle. Proof
  • Let's draw ray BO. In the first case, it was proven that an inscribed angle whose one side passes through the center is equal to half the arc on which it rests. So ∠ ABD = ½ AD.
  • Another angle, the side of which passes through the center — angle CBD, is also equal to half the arc on which it rests. ∠ CBD = ½ CD.
  • Now, if you subtract angle CBD from the large angle ABD, you get angle ABC. ∠ ABC = ∠ ABD − ∠ CBD = ½ AD − ½ CD = ½ AC.

For all three cases, it was proven that the inscribed angle is equal to half the arc on which it rests.

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