In a right triangle ABC, a circle with AB as diameter is drawn to intersect the hypotenuse AC in P. Prove that the tangent at P, bisects the side BC.

Answer

To solve the problem, we will follow a structured approach to prove that the tangent at point P bisects the side BC in triangle ABC.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a right triangle ABC with the right angle at C.
- A circle is drawn with diameter AB, which intersects the hypotenuse AC at point P.

2. **Drawing the Tangent**:
- At point P, draw a tangent line PQ to the circle. We need to prove that this tangent bisects the side BC, meaning we need to show that BQ = QC.

3. **Using the Property of Tangents**:
- From point Q (the external point), we know that the lengths of the tangents drawn from an external point to a circle are equal. Thus, we have:
\[ PQ = BQ \]

4. **Analyzing the Angles**:
- Since AB is the diameter of the circle, by the property of circles, angle APB is 90 degrees (angle subtended by a diameter).
- Therefore, angle BPC is also 90 degrees (since angles on a straight line sum up to 180 degrees).

5. **Examining Triangle BPC**:
- In triangle BPC, we have:
\[ \angle BPC = 90^\circ \]
- This means that:
\[ \angle BPQ + \angle CPQ = 90^\circ \]

6. **Using Triangle Properties**:
- In triangle PBC, we can apply the angle sum property:
\[ \angle PBC + \angle PCB + \angle BPC = 180^\circ \]
- Since \(\angle BPC = 90^\circ\), we have:
\[ \angle PBC + \angle PCB = 90^\circ \]

7. **Establishing Angle Relationships**:
- We know that:
\[ \angle PBC = \angle PBQ \quad \text{and} \quad \angle PCB = \angle PCQ \]
- Therefore, we can substitute:
\[ \angle PBQ + \angle PCQ = 90^\circ \]

8. **Equating Angles**:
- Since we established that both pairs of angles sum to 90 degrees, we can equate them:
\[ \angle BPQ + \angle CPQ = \angle PBC + \angle PCB \]
- This leads us to conclude that:
\[ \angle CPQ = \angle PCB \]

9. **Concluding the Proof**:
- Since \(\angle CPQ = \angle PCB\) and \(PQ = BQ\), by the properties of triangles, we can conclude that:
\[ QC = BQ \]
- Thus, the tangent PQ bisects BC, proving that:
\[ BQ = QC \]

### Final Result:
The tangent at point P bisects the side BC in triangle ABC.

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