1、第十三章轴对称人教版专题训练(五)证明两段线段相等的基本思路1如图如图,在四边形在四边形ABCD中中,ADBC,A90,BDBC,CEBD于点于点E.求证:求证:ADBE.解:证明:解:证明:ADBC,ADBDBC.CEBD,BEC90.A90,ABEC.又又BDBC,ABD ECB.ADBE2如图,在ABC中,点E在AB上,点D在BC上,BDBE,BADBCE,AD与CE相交于点F.求证:EFDF.3如图,在ABC 中,D是AB边上一点,且DCDB.点E在CD的延长线上,且EBCACB.求证:ACEB.4如图所示,12,BDCD.求证:ABAC.解:证明:BDCD,DBCDCB,又12,1D
2、BC2DCB,即ABCACB,ABAC5如图,在ABC中,ACB90,CDAB于点D,AE平分BAC交CD于点F,交BC于点E.求证:CECF.解:证明:ACB90,BBAC90.CDAB,CADACD90,ACDB,AE是BAC的平分线,CAEEAB,EABBCEF,CAEACDCFE,CFECEF,CFCE6如图如图,在在ABC中中,ABAC,EF交交AB于点于点E,交交BC于点于点D,交交AC的延长线于点的延长线于点F,且且BECF.求证:求证:DEDF.7如图,ACAD,线段AB经过线段CD的中点E.求证:BCBD.解:证明:ACAD,E是CD中点,AB垂直平分CD,BCBD8如图,B
3、,E分别是CD,AC的中点,ABCD,DEAC.求证:ACCD.解:证明:连接AD,B是CD的中点,ABCD,AB是线段CD的垂直平分线,ADAC,同理可得ADCD,ACCD9我们把两组邻边相等的四边形叫做“筝形”如图,四边形ABCD是一个筝形,其中ABCB,ADCD.对角线AC,BD相交于点O,OEAB,OFCB,垂足分别是E,F.求证:OEOF.10如图,在四边形ABCD中,ADBC,E为CD的中点,连接AE,BE,BEAE,延长AE交BC的延长线于点F.求证:(1)FCAD;(2)ABBCAD.(2)ADE FCE,AEEF.AEBE,BE是线段AF的垂直平分线,ABBFBCCF.ADCF,ABBCAD11如图,在ABC中,ABAC,A36,BD是B的平分线,交AC于点D,点E是AB中点,ED交BC的延长线于点F.求证:ABCF.证明:如图,连接AF,ABAC,BAC36,ABCACB72.BD平分ABC,1236,1BAD,DADB.AEBE,FEAB,即FE是AB的垂直平分线,FAFB,FABABC72,3FABBAC36.ACB3AFC,AFCACB336,3AFC,ACCF,ABCF