Figure 1
Example 1 (20 1 1 Bijie, Guizhou) As shown in figure 1, after a circular paper with a radius of 2 cm is folded in half, the arc just passes through the center o, and the length of the crease AB is
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Answer? 2 cm B？ 3 cm C？ 23 cm D？ 25 cm
By analyzing the chord of vertical connection radius, a right triangle is constructed in the figure, and the length of AD is calculated according to Pythagorean theorem, and then the length of AB is calculated according to vertical diameter theorem.
OD⊥AB is in Zone D, connecting OA. From origami, OD= 12OA= 1 cm, at Rt△OAD, AD=3 cm, ∫od⊥ab, ∴ AB=2AD=23 cm. Therefore,
It is the key to solve the problem that OD= 12OA= 1 is obtained by folding.
Figure 2
Example 2 (20 1 1 Sanming, Fujian) As shown in Figure 2, on a square piece of paper, ABCD, E and F are the midpoint of AD and BC respectively. Fold the figure along a straight line passing through point B, so that point C falls on EF, and the falling point is n, the crease intersects the CD edge at point M, BM intersects EF at point P, and then unfolds. Then the following conclusions are drawn: ① cm = ② ∠ ABN = 30; ③AB2 = 3c m2； ④ △PMN is an equilateral triangle.
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Answer? 1 B？ Two c's and three d's? four
Analysis ∵ E and F are the midpoint of AD and BC, ∴ef∨ab respectively, from which we can get EF ∵ BC. ∫BF = 12bn, bfn, sin in right triangle. ∴ BCCM=3, that is, BC2=3CM2, and the answer ③ is correct. It is also CM=32BC=32DC, DM= 1-32DC, and the answer ① cm = DM is incorrect. It is also ∠ CMB = ∠ NMB = 60.
It is the key to solve this problem that the line segments and angles are equal before and after folding.
Figure 3
Example 3 (Neijiang, Sichuan, 20 1 1) As shown in Figure 3, in the rectangular coordinate system, the edge OA of the rectangular ABCO is on the X axis, the edge OC is on the Y axis, and the coordinate of point B is (1, 3). Fold the rectangle along the diagonal AC, the point B falls at the position of point D, and the AD intersects the Y axis at point E..
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Answer? -45 125 B？ -25 135 C？ - 12， 135 D？ -35, 125
By analyzing the nature of folding, it is determined that △AEC is an isosceles triangle, and the length of AE is calculated by Pythagorean theorem, then the length of EC is calculated, and then the coordinates of point D are calculated according to similarity or trigonometric function.
Analysis is DM⊥x axis, vertical foot is M, and folding can get ∠CAB=∠CAD, from ab∨oc, ∠CAB=∠ACO, knowing ∠ACO=∠CAD, ∴ EA=EC. Let D(x, y), sin ∠ eao = oeae = dmad, that is, 4353=y3, y= 125, tan∠EAO=OEOA=DMAM, that is, 431=125.
Comment on the condition that the bisector and parallel line form an isosceles triangle: when calculating the coordinates of the midpoint in the rectangular coordinate system, the definition of trigonometric function or the proportional formula similar to the triangle is often used to solve it.
Example 4 (20 1 1 Zunyi, Guizhou) Fold rectangular ABCD paper as shown in Figure 4, so that point A and point E overlap, point C and point F overlap (both points E and F are on BD), and the creases are BH and DG respectively.
Figure 4
(1) Verification: △ bhe △ DGF;
(2) If AB = 6 cm and BC = 8 cm, find the length of the line segment FG.
Analysis (1)∫ quadrilateral ABCD is a rectangle, ∴ AB=CD, ∠ A = ∠ C = 90, ∠ABD=∠BE=DF, and ∠ 1 = \ is obtained by folding.
(2) ∵ quadrilateral ABCD is a rectangle, AB=6 cm, BC=8 cm. According to Pythagorean Theorem, BD =10cm, ∴ BF =10-6 = 4cm. Let FG=x, BG=8-x, at RT △ bg.
Comments: Grasping the shape, size and position changes of the graphics before and after folding, and the corresponding edges and angles are equal, the problem can be solved smoothly.
Example 5 (20 1 1 Shenzhen, Guangdong) As shown in Figure 5, a rectangular piece of paper ABCD, in which AD = 8 cm and AB = 6 cm, is first folded in half along the diagonal BD, with point C at point C' and point BC' at point G. 。
(1) verification: ag = c' g;
(2) Fold it again, as shown in Figure 6, so that point D coincides with point A, and a crease EN is obtained. EN intersects with AD at point M, and the length of EM is found.
(1), from the nature of isosceles triangle, we can know that AD=BC=BC', ∠ 1=∠2=∠3, GB=GD, so Ag = c'. (2) In RT △ C ′ DG, the length of C ′ g is obtained by Pythagorean theorem, and then the length of em is obtained.
The analysis (1) is shown in Figure 5. From the symmetry of the graph, BC=BC', ∠ 1=∠2.
∫ quadrilateral ABCD is a rectangle, ∴ AD=BC, ad∨BC, ∴ ∠2=∠3, so ∠ 1=∠3, GB=GD.
And AD=BC', ∴ ag = c' g.
Figure 5
Figure 6
Figure 7
(2) as shown in fig. 6, if Ag = X, then C' G = X, DG = 8-X, DM= 12AD=4 cm. At Rt△C'DG, ∠ DC' G = 90, C' D = CD = 6cm, ∴.