Lemma: Let C be a circle with center O, and let the acute angle XYZ be inscribed in circle C. Then the measure of angle XYZ is half the measure of angle XOZ.

Click here to verify the theorem using GSP.

Proof of Lemma:

Let C be a circle with center O, and let <XYZ be an acute angle that is inscribed in circle C.

Construct segments XZ and OY.

Segments OX, OZ, and OY are radii of the circle, so they are congruent. Therefore, triangles OXZ, OYZ, and OXY are isosceles triangles. Since base angles of an isosceles triangle are congruent, we have

m<OZY = m<OYZ

m<OZX = m<OXZ

m<OXY = m<OYX.

Label the angles as follows:

a = m<OZY = m<OYZ

b = m<OZX = m<OXZ

c = m<OXY = m<OYX

d = m<XOZ.

First consider triangle OXZ. Using the fact that the sum of the measures of the interior angles of a triangle is 180 degrees, we have

b + b + d = 180
2b + d = 180
d = 180 - 2b.

Considering triangle XYZ, we have

(b + c) + (c + a) + (a + b) = 180
2a + 2b + 2c = 180
2a + 2c = 180 - 2b.

Since d = 180 - 2b and 2a + 2c = 180 - 2b, we can now write

d = 2a + 2c
1/2 d = a + c
1/2 m<XOZ = m<OYZ + m<OYX = m<XYZ.

Therefore, the measure of the inscribed angle (<XYZ) is equal to half the measure of the central angle (<XOZ).