病例1 如图1,已知AB∥CD,求证:∠ABE+ ∠BED+∠EDC=360°．证明:过点E作AB、CD的平行线EF,X 因为EF∥AB,所以∠ABE+∠BEF=180°．因为EF∥CD,所以∠FED+∠EDC=180°. 而∠BEF+∠FED=∠BED,所以∠ABE+∠BED+ ∠EDC=360°．病因:违背了平行公理,过直线外一点有且只有一条直线和已知直线平行,即过点E不能作一条直线既与AB平行,又与CD平行, 只能先作出和其中一条直线平行的直线,然后再去证明它也与另一 Case 1 As shown in Figure 1, known AB∥CD, verification: ∠ABE+ ∠BED+∠EDC=360°. Prove that: over the point E for AB, CD parallel line EF, X because EF ∥ AB, so ∠ ABE + ∠ BEF = 180 °. Since EF∥CD, ∠FED+∠EDC=180°. And ∠BEF+∠FED=∠BED, so ∠ABE+∠BED+ ∠EDC=360°. Etiology: contrary to the parallel axioms, there is only one straight line and a straight line parallel to the known straight line beyond the straight line, that is, the cross point E can not be a straight line both parallel to the AB and parallel to the CD, can only be made parallel to one of the straight lines first Straight line and then go and prove it also with another