Point p, point q and point w are the midpoint of FM, MN and NF respectively.
∴pq= 12fn,qw= 12fm,pw= 12mn.
∴FNPQ=FMQW=MNPW=2.
∴△FMN∽△QWP.
(2) solution: ①∵ quadrilateral ABCD is rectangular,
∴∠DAB=∠B=∠C=∠D=∠90,AD=BC=4cm,DC=AB=6cm。
From the title, it can be seen that DM = BN = 1× t = t (cm).
So AM=.4? t .,AN=6-t。
∴MN2=AM2+AN2
=(4-t)2+(6-t)2
=2t2-20t+52
=2(t-5)2+2。
∵2>0,
When t=5, MN2 is the smallest, and the minimum value is 2.
When t=5, MN takes the minimum value and the minimum value is 2.
②∫△FMN∽△QWP,
∴∠PQW=∠NFM,∠QWP=∠FMN,∠WPQ=∠MNF.
1. When ∠ PQW = 90, ∠ NFM = 90.
The intersection n is NE⊥DC and the vertical leg is e, as shown in Figure 2.
∠∠MDF =∠MFN =∠FEN = 90,
∴∠DFM=90 -∠EFN=∠ENF。
∴△MDF∽△FEN.
∴DMEF=DFEN.
DM = t,DF=2,EF=CF-CE=6-2-t=4-t,EN=BC=4,
∴t4? t=24。
Solution: t = 43.
It is verified that t=43 is the solution of the equation, which accords with the meaning of the question.
Ⅱ. When ∠ PWQ = 90, ∠ NMF = 90.
Similarly: △ MDF ∽△ Nam.
There is DFAM=DMAN ..
DF = 2,DM=t,AM=4-t,AN=6-t,
∴24? t=t6? t.
Finishing: T2-6t+ 12 = 0.
∵(-6)2-4× 1× 12=- 12