It is found that with P as the vertex, point M and point N are isosceles right angles △PMN on both sides of △ A respectively.

Analysis: Assume that △PMN has been done, as shown in the figure. Make AN isosceles right angle △PBC with P as the vertex so that BC is on the ray An, then △PMN∽△PBC, ∠PMN=∠PCB, P, N, C, M * * circle, ∠PCM=∠PNM=Rt∠.

Practice: 1, the intersection point p is that PB is perpendicular to the ∠A side, and the vertical foot is b;

2. Intercept BC=BP at the edge of ∠A and connect PC;

3. The point C is CM⊥PC ∠A, and the other side is connected to PM at point M;

4. Make the midpoint of PM O;

5. Take O as the center, PO as the radius, and make an arc ray AC at point N to connect MN and PN, then △PMN is the demand.

It is proved that △PBC is an isosceles right triangle and △PMN is a right triangle.

Then from ∠PCM =∠PNM = Rt∞, the four-point * * circle of p, n, c and m is obtained.

So ∠ PMN = ∠ PCB = 45,

It can be concluded that △PMN is an isosceles right triangle.

Discussion: This question is put forward according to the given conditions. Because point C can be on both sides of line segment PB, this problem can have two solutions, but if there is only one intersection between the vertical line of PC passing through point C and ∠A, this problem has only one solution, and if there is no intersection, this problem has no solution. Moreover, if point P is set as a right-angled vertex, the solution will be different, leaving it to interested friends to discuss.