As shown in figure 1, in a square, the diagonal line intersects a point and the bisector intersects a point.

(1) Verification:;

(2) The point starts from the point and moves to the point along the line segment (not coincident with the point). At the same time, the point starts from the point and moves along the extension line of, and the moving speed of the point is the same as that of the point. When one moving point stops moving, the other moving point also stops moving. As shown in figure 2, divide equally, cross at points, do at points, and stand on their feet. Please guess and prove your guess.

(3) Under the condition of (2), when, when and how long.

It is proved that (1) as shown in figure 1, the intersection point F is FM⊥AB of point M, and in the square ABCD, it is ac⊥bd∴AE = AC, ∠ Abd = ∠ CBD = 45 of point E.

∫AF ∠BAC,

∴EF=MF，

AF = AF，

∴Rt△AMF≌Rt△AEF，

∴AE=AM，

∠∠MFB =∠ABF = 45，

∴MF=MB,MB=EF，

∴EF+ AC=MB+AE=MB+AM=AB。

(2) The quantitative relationship between E1f1A 1C 1 f1a1c65438 = AB.

Proof: As shown in Figure 2, connect F 1C 1 and pass through F 1 to make F 1P⊥A 1B at point P and F 1Q⊥BC at point Q,

∫a 1f 1 divide equally ∠BA 1C 1, ∴ e1f1= pf1; Similarly QF 1=PF 1, ∴ e 1f 1 = pf 1,

∵a 1f 1 = a 1f 1，∴rt△a 1e 1f 1≌rt△a 1f 1，

∴A 1E 1=A 1P，

Similarly rt △ qf1c1≌ rt △ e1f1,

∴C 1Q=C 1E 1，

Meaning: A 1A=C 1C,

∴a 1b+bc 1=ab+a 1a+bc-c 1c=ab+bc=2ab，

∫PB = pf 1 = qf 1 = QB，

∴a 1b+bc 1=a 1p+pb+qb+c 1q=a 1p+c 1q+2e 1f 1，

Namely 2ab = a1e1+c1e1+2e1f1= a1e1,

∴e 1f 1+ a 1c 1 = ab。

③ Let PB=x, then QB=x,

∫a 1e 1 = 3，QC 1=C 1E 1=2，

Rt△A 1BC 1，a 1 B2+BC 12 = a 1c 12，

That is, (3+x)2+(2+x)2=52,

∴x 1= 1, x2=-6 (omitted),

∴PB= 1，

∴E 1F 1= 1，

∫a 1c 1 = 5，

The conclusion from (2): e1f1+a1c1= ab,

Calculate the answer yourself.