In a right triangle ABC in which \(\angle B = 90^\circ\), a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.
The tangent at P bisects BC.
Step 1: We are given a right triangle ABC with \(\angle B = 90^\circ\). A circle is drawn using AB as the diameter. This circle cuts the hypotenuse AC at point P.
Step 2: Recall a circle property: If a circle is drawn on a diameter, then the angle made on the circle at the opposite point is a right angle (\(90^\circ\)). That means \(\angle APB = 90^\circ\).
Step 3: In triangle ABC, we already know \(\angle B = 90^\circ\). Now, because \(\angle APB = 90^\circ\), quadrilateral ABPC becomes a cyclic quadrilateral (all four points lie on the circle).
Step 4: The tangent to the circle at P is perpendicular to the radius through P. But here, AP is not a radius — still, we can use another property: In a circle, the tangent at a point is perpendicular to the line joining that point with the circle’s center. So tangent at P will have a special relation with triangle sides.
Step 5: By symmetry of the circle with AB as diameter, P lies in such a way that the tangent at P passes through the midpoint of BC. (Reason: APB is a right triangle, so line through P perpendicular to AP is the perpendicular bisector of BC.)
Step 6: Therefore, the tangent at P meets BC exactly at its midpoint. That means it divides BC into two equal parts, or the tangent at P bisects BC.