Final Answer:
The tangent to the circle at point P, where a circle with AB as diameter intersects the hypotenuse AC in a right triangle ABC, bisects BC. This is proven by the property that a line tangent to a circle is perpendicular to the radius at the point of contact, resulting in equal segments formed by the tangent on the circle.
Step-by-step explanation:
Consider the right triangle ABC with a 90° angle at B. Let AB be the diameter of the circle passing through points A, B, and C. The point of intersection between the circle and the hypotenuse AC is denoted as P. As per the property of tangents to a circle, a tangent drawn from an external point to the circle is perpendicular to the radius at the point of contact. Therefore, the tangent at point P is perpendicular to the diameter AB.
Since AB is a diameter, it passes through the center of the circle. This implies that BP and CP are radii of the circle, and they are equal in length because they both extend to points of tangency. As the tangent from point P is perpendicular to the diameter AB, it divides the diameter into two equal parts, BP and CP. Therefore, in the right triangle ABC, the tangent to the circle at point P bisects the side BC, dividing it into two equal segments, fulfilling the condition of being a midpoint. Hence, the tangent indeed bisects BC in the given triangle.