Theorem One: Proving AP = PB

Given Information

• OP is perpendicular to AB
• O is the center of the circle
• Required to prove: AP = PB

Construction Steps

1. Construct lines OA and OB.
   • This forms two triangles:
     • Triangle AOP
     • Triangle OBP

Triangle Analysis

• We need to prove that AP is equal to PB using congruency of triangles AOP and OBP.

Identifying Congruent Parts

1. Angles

   • Angle P1 (in triangle AOP) is equal to Angle P2 (in triangle OBP)
   • Reason: Given that OP is perpendicular to AB.

2. Common Side

   • OP is a common side for both triangles.
   • Therefore, OP = OP (common side).

3. Radii

   • OA = OB (both are radii of the circle).
   • Therefore, OA = OB.

Conclusion of Congruency

• With the above identifications:
  • AP = PB is proven by the congruence of triangles AOP and OBP.
  • Reasoning:
    • Right Angle: Each triangle has a right angle (90 degrees) at P.
    • Hypotenuse: OA and OB are the hypotenuses (both radii).
    • Common Side: OP is the common side.

Congruence Statement

• Triangle AOP is congruent to Triangle OBP by the Right Angle - Hypotenuse - Side (RHS) theorem.

Final Statement

• Therefore, AP = PB because Triangle AOP is congruent to Triangle OBP.