Solution: Let the side length of square ABCD be 1, AP pass through BC in E, BP pass through AC in O, and connect OD.

Because ∠ABC=RT∠ AB=BC P is the heart of △ABC,

∴BO split ∠ABC AE split ∠BAC

∴AO=OC Then O is the center of the square ABCD.

∴B, p, o and d are on the same straight line,

∠DAP=∠DAC+∠CAE=45 +∠BAC/2

∠DPA=∠ABD+∠BAE=45 +∠BAC/2

∴∠DAP=∠DPA

∴DP=DA