As shown in the figure: in square ABCD, AC = 10, P is any point on AB, PE ⊥ AC is in E, PF ⊥ BD is in F, then PE + PF=______ It can be summed up in one sentence: the sum of the distances from any point on the side of a square to two diagonals is equal to______ ．

As shown in the figure: in square ABCD, AC = 10, P is any point on AB, PE ⊥ AC is in E, PF ⊥ BD is in F, then PE + PF=______ It can be summed up in one sentence: the sum of the distances from any point on the side of a square to two diagonals is equal to______ ．

∨ PE ⊥ AC, BD ⊥ AC ∨ PE ∥ Bo, ∨ ape ∨ ABO, ∨ pebo = apab, the same theorem can be proved: pfao = bpab, ∨ apab + bpab = pebo + pfao = ABAB = 1, ∨ Ao = Bo, ∨ PE + pf = Ao = Bo, ∨ AC = 10, ∨ Ao = 5, so PE + pf = 5, so we can use a sentence to summarize: any point on the edge of a square to two pairs of