Question

67. In figure, a triangle ABC with ZB 90° is shown. Taking AB as diameter, a circle has been drawn intersecting AC at point P. Prove that the tangent drawn at point P bisects BC.

Answer 1

The tangent to the circle at P bisects BC.

Answer 2

Explanation of the solution:

1. Step 1: In triangle ABC with ∠B = 90° and with AB as a diameter, Thale’s theorem implies that ∠APB = 90°.

2. Step 2: Draw the tangent to the circle at P. By the tangent–chord angle theorem, the angle between the tangent at P and the chord PA equals the inscribed angle in the alternate segment, namely ∠PBA.

3. Step 3: Using these equal angles and the fact that Q (where the tangent meets BC) lies on the line BC, one shows that the triangles QBP and QCP have two equal angles. Thus, by AA (angle–angle) they are congruent.

4. Step 4: Congruence of these triangles immediately gives QB = QC, i.e. the tangent at P bisects BC.