已知:AB=CD,角A=角D,求证:角B=角CA ___________D/.\/.\ B -----------------------C
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![已知:AB=CD,角A=角D,求证:角B=角CA ___________D/.\/.\ B -----------------------C](/uploads/image/z/13378087-55-7.jpg?t=%E5%B7%B2%E7%9F%A5%3AAB%3DCD%2C%E8%A7%92A%3D%E8%A7%92D%2C%E6%B1%82%E8%AF%81%3A%E8%A7%92B%3D%E8%A7%92CA+___________D%2F.%5C%2F.%5C+B+-----------------------C)
已知:AB=CD,角A=角D,求证:角B=角CA ___________D/.\/.\ B -----------------------C
已知:AB=CD,角A=角D,求证:角B=角C
A ___________D
/.\
/.\
B -----------------------C
已知:AB=CD,角A=角D,求证:角B=角CA ___________D/.\/.\ B -----------------------C
连接AC、BD,则根据边角边对应相等得三角形ABD全等于三角形DCA,进而AC=BD
然后根据边边边对应相等得三角形ABC全等于三角形DCB,进而角ABC=角DCB
连结AC,BD,因为AB=CD,AD=AD,角A=角D,所以三角形ABD、DCA全等,所以AC=BD,因为AB=DC,AC=BD,BC=BC,所以三角形ABC、DCB全等,所以角B=角C