Two circles touch each other externally at P. AB is a direct common tangent to the two circles, A and B are points of contact, and ∠PAB=40∘. The measure of ∠ABP is:

For two circles touching externally at P, AB is a direct common tangent with A and B as points of contact. The radius to the point of tangency is perpendicular to the tangent, so OA is perpendicular to AB and OB is perpendicular to AB. This means OA and OB are parallel. Given ∠PAB = 40°, the line PA makes a 40° angle with AB. Since OA is perpendicular to AB, ∠PAO = 90° - 40° = 50°. Triangle PAO is isosceles (OA = OP as radii), so ∠ABP = ∠PAO = 50°. Alternatively, using properties of tangents and radii, the angle between the tangent and chord equals the angle in the alternate segment.