Theorem:

Given a circle with diameter AB, and given a third point C on the circle, the angle ACB is a right angle.

We are going to prove the above-statesd theorem using this figure.

Let D be the center of the above circle. AB is shown to be the diameter of the circle so the center D lies on the diameter AB. By definition of the radius of a circle, we can conclude that segments AD = BD = CD.

Therefore, are isosceles triangles.

By the isosceles triangle theorem, the Angle DAC and Angle DCA are equal as well as Angle DBC and Angle DCB are equal.

Let a = Angle DBC = Angle DCB. Let b = Angle DAC = Angle DCA.

The sum
of the angles of triangle ABC should

add up to 180 degrees from the __Angle Sum Theorem__.

So, therefore a+(a+b) + b = 180

2a + 2b = 180

a+b = 90

So angle ACB is a right triangle equaling 90 degrees.

Therefore, given a circle with diameter AB, and given a third point C on the circle, the angle ACB is a right angle.