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.
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.