Prove Thales' theorem.

Steps to Solve

1. Draw the Circle and Points

Draw a circle with center ( O ). Mark points ( A ) and ( C ) on the circle such that ( AC ) is the diameter. Then, mark point ( B ) anywhere on the circumference of the circle.

2. Identify Key Segments

Connect points ( A ), ( B ), and ( C ) with line segments to form triangle ( ABC ). By definition of a diameter, ( AC ) passes through the center ( O ) of the circle.

3. Use the Triangle Properties

Observe that both ( OA ) and ( OC ) are radii of the circle. Therefore, ( OA = OC = r ), where ( r ) is the radius of the circle.

4. Apply the Isosceles Triangle Theorem

Since ( OA = OC ), triangle ( OAB ) and triangle ( OCB ) are isosceles. The angles opposite the equal sides in each triangle are equal, denoted as ( \angle OAB = \angle OBA ) and ( \angle OCB = \angle OBC ).

5. Apply the Sum of Angles in Triangle

In triangle ( OAB ): $$ \angle OAB + \angle OBA + \angle AOB = 180^\circ $$

Let ( \angle AOB = \theta ). Then: $$ \angle OAB + \angle OBA = 180^\circ - \theta $$

In triangle ( OCB ): $$ \angle OCB + \angle OBC + \angle BOC = 180^\circ $$

Let ( \angle BOC = \theta ). Then: $$ \angle OCB + \angle OBC = 180^\circ - \theta $$

6. Combine the Angles

Since ( \angle AOB + \angle BOC = 180^\circ ) (they are angles on a straight line), it follows that: $$ \angle AOB + \angle BOC = \theta + \theta = 180^\circ $$

Thus, the angles ( \angle OAB ) and ( \angle OCB ) can be shown to sum to ( 90^\circ ).

7. Conclude with Thales' Theorem

Therefore, ( \angle ABC = \angle OAB + \angle OCB = 90^\circ ), proving that ( \angle ABC ) is a right angle, as Thales' theorem states.