Given: Two circles with centers A and B touch each other at point P. To Prove: Points A, P, and B lie on the same straight line. Construction:Join A to P and B to P. Proof: The circles with centers A and B touch each other at point P. ∴ There exists a common tangent at point P to both circles. Let ST be the common tangent that touches both circles at point P. Since ST is a tangent to the circle centered at A, and AP is the radius to the point of contact, ∴ AP ⊥ ST Similarly, since ST is also a tangent to the circle centered at B, and BP is the radius to the point of contact, ∴ BP ⊥ ST ∴ Both AP and BP are perpendicular to the same line ST at point P. ∴ AP and BP lie along the same straight line. Hence, A, P, and B are collinear. (Proved)