The simplest way to solve this problem is to add equations by Pythagorean theorem.

Solution: ∵ABCD is rectangular.

∴∠A=∠D=∠ABC=∠DCB

∴△ APB and△ PDC are right triangles.

∫△PHF is a right triangle.

∴△ APB, △ PDC and △ PHF are all right triangles.

Let AP be x, then PD =10-X.

∴BP？ =X？ +4? ，PC？ =( 10-X)？ +4?

BC? = 10? = 100

△ PBC is a right triangle.

∴X？ +4? +( 10-X)？ +4? = 100

∴X？ - 10X+ 16=0

∴X？ - 10X+25-25+ 16=0

∴(X-5)？ =9

∴X-5= 3

X = 8 or 2

A: Yes, the AP length is 8 or 2.