Proof:

Step 1: Draw the diagram and label the given information

- Draw a circle with centre O and diameter AB.
- Draw the semicircle APB with P on it.
- Draw tangents to the circle at points A and B.
- Draw perpendiculars AH and BK from A and B respectively to the tangent at P.
- Label AB as the diameter of the circle.

Step 2: Show that triangles AHP and BKP are similar

- Both triangles have a right angle at P.
- Angle AHP and angle BKP are both angles formed between a tangent and a radius of the circle, so they are equal.
- Therefore, by angle-angle similarity, triangles AHP and BKP are similar.

Step 3: Use the similarity of triangles AHP and BKP to find the length of AH and BK

- Let x be the length of AP.
- Then, by the Pythagorean theorem, AH^2 = AP^2 - PH^2 = x^2 - OP^2.
- Similarly, BK^2 = BP^2 - PK^2 = (AB - x)^2 - OP^2.
- Since triangles AHP and BKP are similar, we can set up the proportion AH/BK = HP/KP.
- Substituting the expressions for AH and BK, we get (x^2 - OP^2)/((AB - x)^2 - OP^2) = HP/KP.
- Cross-multiplying, we get HP*(AB - x)^2 = KP*x^2.
- Simplifying, we get HP/AB = x/(AH + BK).

Step 4: Use the fact that HP/AB = 1/2 to find AH*BK

- Since P is on the semicircle, angle APB is a right angle.
- Therefore, angle AHP + angle BKP = 90 degrees.
- Using the fact that HP/AB = 1/2, we can rewrite the equation from Step 3 as 2x/(AH + BK) = 1.
- Solving for AH + BK, we get AH + BK = 2x = AB.
- Therefore, AH*BK = (AH + BK)*HK = AB*HK = AB/2*AB/2 = AB^2/4.

Step 5: Conclusion

- We have shown that AH*BK = AB^2/4.
- Therefore, AH*BK = AB*AB/4 = AB^2/4.
- Hence, AH*BK=AB.