1. Point p is on AC:

1, the intersection point f is an alternating vertical line, and the vertical foot H-EFHC is a rectangle-HF = ce;

2. the intersection point p is the perpendicular of EF, and the vertical foot is g, because PEQF is a diamond-triangle EPF is an isosceles triangle-ge = gf, so CP = HP.

3. It is known that AP = 3t-PC = 6-3t-Ah = 6-2 (6-3t); CE = HF = 4/3t；

4. Substitute the similarity (HF is parallel to BC) -HF/8 = AH/6- into the quantitative relationship in (3) to get the value of t.

Second, the point P is on CB: Because the symmetry point is on BC, the three-point * * * line cannot form a diamond.

3. point p is on BA:

1, because PEQF is a diamond-triangle EPF is an isosceles triangle-EP = FP-angle PEF= angle PFE；; ;

2. It is easy to know that the angle BEF=90 degrees, so the angle PBE= the angle PEB- the midpoint of the point P BF;

3. The movement time of point P is 2(AC)+2(BC)+T(BP)=t,-t = t-4; So BP = (t-4) * 5-BF = 2 (t-4) * 5;

4、EC = 4/3t-BE = 8-4/3t；

5. Substitute the similarity (EF is parallel to AC)-be/8 = BF/ 10- into the quantitative relationship in Formula (3) and Formula (4), and the value of t can be obtained.

I wish you good grades!