基本模式如图1,若点E是线段AB上的一点,且∠CAE=∠EBD=∠CED=90°,则CAE∽EBD.(当CE=DE时,CAE≌EBD.)(证明略.)基本模式的变形如图2,把EBD平移,使得点E移至点A处,此时CE和AD的交于点H,同时保证∠CHD=90°.这里同样可以利用基本模式的方法证明出两个三角形相似的结论.图3是图2在动态的情况下的特殊情况,即当点D作为动点,当该点和点F重合后的情形.此时在正方形内,CAE和EBD都变成了等腰直角三角形,如果点C和点E都作为动点的话,最后 The basic mode is shown in Figure 1. If point E is a point on line AB and ∠CAE = ∠EBD = ∠CED = 90 °, then ∽CAE∽EBD. (When CE = DE, CAE≌EBD.) (Prove slightly.) Deformation of the basic model shown in Figure 2,  EBD translation, so that point E moved to point A, at this time CE and AD at point H, while ensuring ∠ CHD = 90 °. The basic mode method proves the conclusion that the two triangles are similar.Figure 3 is the special case of Fig. 2 under the dynamic condition, that is, when the point D is the moving point and the point coincides with the point F. At this time, , CAE and EBD have become isosceles right triangle, if the point C and point E are used as moving point, the last