(Mianyang, 2011) as shown in the figure, fold a rectangular piece of paper ABCD with a length of 8cm and a width of 4cm so that point a and point C coincide, and the length of crease EF is equal to 2525cm

(Mianyang, 2011) as shown in the figure, fold a rectangular piece of paper ABCD with a length of 8cm and a width of 4cm so that point a and point C coincide, and the length of crease EF is equal to 2525cm

Connect AC, intersect with EF at o point, ∵ e point is on AB, f is on CD, because points a and C coincide, EF is crease, ∵ Ao = Co, EF ⊥ AC, ∵ AB = 8, BC = 4, ∵ AC = 45, ∵ AE = CE, ∵ EAO = ∵ Eco, ∵ OEC ∽ BCA, ∵ OE: BC = OC: Ba, ∵ OE = 5, ∵ EF = 2oe = 25

As shown in the figure, fold the rectangular piece of paper ABCD along EF, so that point d coincides with point B, and point C falls on the position of point C ′. (1) try to explain that △ bef is an isosceles triangle; (2) are there two figures in the figure which are symmetrical in the center? If it exists, please indicate which two figures are (no need to explain the reason, solid line and dotted line in the figure are treated the same); (3) if AB = 4, ad = 8, find the length of crease EF

(1) ∵ ed ‖ FC, ∵ def = ∠ BFE, according to the folding invariance, ∵ def = ∠ bef is obtained, so ∵ bef = ∠ BFE. △ bef is isosceles triangle; (2) trapezoid CFED and trapezoid aefb are centrosymmetric graphs; (3) eg ⊥ BF is made in G. let AE = x, then ed = 8-x, according to the folding invariance, be = ed = 8-x. in RT △ Abe, X2 + 42 = (8-x) 2, the solution is x = 3. So be = 8-3 = 5, and be = BF Then GF = 5-3 = 2. EF = EG2 + GF2 = 25

It is known that for a rectangular piece of paper ABCD (AD > AC) as shown in the figure, fold the paper once so that the points a and C coincide, and then unfold. The crease EF intersects the ad edge at e and BC edge
Is there a point P on the line AC such that 2ae & # 178; = AC × AP? If so, please specify the position of point P and prove that (it has been proved that afce is a diamond,

Solution: points a and C are symmetric with respect to EF, then EA = EC, ∠ EAC = ∠ ECA. Make the vertical bisector of AE, intersect AC at m, connect em, then Ma = me, ∠ mea = ∠ EAM. And ∠ EAM = ∠ CAE, then ⊿ EAM ≛ CAE, AE / AC = am / AE, AE & # 178; = AC × am, 2ae & # 178; = AC × (2am). Intercept MP = am, then AP = 2am. We can get: 2

The edge length of rectangular paper ABCD is ab = 4, ad = 2. Fold the rectangular paper along EF so that point a and point C coincide. After folding, color it on one side (as shown in the figure), the area of the colored part is ()
A. 8B. 112C. 4D. 52

In RT △ GFC, there are fc2-cg2 = fg2, fc2-22 = (4-fc) 2, the solution is FC = 2.5, and the area of 〈 shadow part is ab · ad-12fc · ad = 112, so B