Therefore, PW=MN/2, PQ=NF/2 and QW=MF/2.

So pw: Mn = pq: nf = qw: MF

Easy to get: △FMN is similar to △QWP.

2. According to (1), when △PQW is a right triangle, △FMN is a right triangle. Therefore, when △FMN is a right triangle, the problem is transformed into finding the value of x.

Stop moving because of N to A. Therefore, 0 < = t < = 6.

NF^2=(4-x)^2+ 16

MF^2=x^2+4

MN^2=(4-x)^2+(6-x)^2

① When ∠MFN=90 degrees, NF 2+MF 2 = Mn 2.

That is, (4-x) 2+16+x2+4 = (4-x) 2+(6-x) 2.

x=4/3

② When ∠FMN=90 degrees, Mn 2+MF 2 = NF 2.

Namely, (4-x) 2+(6-x) 2+x2+4 = (4-x) 2+16.

X has no solution

③ When ∠MNF=90 degrees, NF 2+Mn 2 = MF 2.

That is, (4-x) 2+16+(4-x) 2+(6-x) 2 = x2+4.

X=4，x= 10 (s)

To sum up, when x=4/3 or 4, △PQW is a right triangle.

When 0 =

3、MN^2=(4-x)^2+(6-x)^2

=2*(x^2- 10x+26)

=2*(x^2- 10x+25+ 1)

=2*[(x-5)^2+ 1]

When x=5, take the minimum value, Mn 2 = 2, MN=√2.