This question mainly examines the knowledge of fillet ruler, and the key to solving the problem is to make clear that the fillet ruler opposite to the diameter is a right angle.

In the first conclusion, AB is the diameter, so ∠ ACB = 90, that is to say, AC is perpendicular to BF, but it cannot be concluded that AC divides BF equally, so it is wrong. In the second conclusion, FP is bisected only when it passes through the center of the circle, so AC bisection ∠BAF cannot be proved.

Firstly, four * * circles of D, P, C and F are proved, and then △AMP∽△FCP is used to draw the conclusion that the circumferential angle of the diameter is a right angle.

Prove: ①∫AB is the diameter,

∴∠ACB=90，

∴AC is perpendicular to BF, but it cannot be concluded that AC divides BF equally.

So 1 is wrong, and 2 is only divided equally when FP passes through the center of the circle, so if FP does not pass through the center of the circle, AC division ∠BAF cannot be proved.

So 2 is wrong, as shown in Figure 3: This is the detailed answer/exercise/math /79949 1. Point P is in a semicircle with a diameter of AB, connecting AP and BP, and extending the semicircle to points C and D respectively, connecting AD and BC, and extending the intersection point to point F as a straight line PF. The following statement must be correct. The topic is still very difficult, but after reading the above ideas and answers, I believe you will understand. If you find it useful, I hope to adopt it ~ Come on, I wish you progress in your study!