八年级数学下册第十八章四边形18.1.2平行四边形的判定(二)课件(新版)新人教版.ppt

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1、,核心目标,掌握平行四边形的判定方法,并会简单运用,课前预习,相等,2.在四边形ABCD中,已知ABCD,请补充一个条件_,使得四边形ABCD是平行四边形,1.一组对边平行且_的四边形是平行四边形,ABCD,课堂导学,知识点:一组对边平行且相等的四边形是平行四边形.,【例题】(2015黄冈)已知:如下图,在四边形ABCD中,ABCD,E,F为对角线AC上两点, 且AECF,DFBE.求证:四边形ABCD为平行 四边形,【解析】首先证明AEBCFD可得ABCD,再由条件ABCD可得四边形ABCD为平行四边形,课堂导学,【答案】证明:ABCD,BACDCA, DFBE,BECDFA,AEBCFD.

2、 在AEB和CFD中AEBCFD(ASA),ABCD,ABCD, 四边形ABCD为平行四边形 【点拔】此题主要考查了平行四边形的判定,关键是掌握一组对边平行且相等的四边形是平行四边形.,课堂导学,对点训练,1.如下图,在ABCD中,AECF,求证:四边形 DEBF是平行四边形,在ABCD中, 则ABCD,ABCD, AECF,ABAECDCF, BEDF,BEDF, 四边形DEBF是平行四边形,课堂导学,(1)证明:四边形ABCD为平行四边形,ADBC,ADBC,ADECBF.又AEBD于E,CFBD于F,AEDCFB90,ADECBF;,课堂导学,(2)由(1)知,ADECBF,则AECF,

3、AEBD,CFBD,AEFC,四边形CEAF是平行四边形,课堂导学,3.已知如下图,O为平行四边形ABCD的对角线AC 的中点,EF经过点O,且与AB交于E,与CD 交于F. 求证:四边形AECF是平行四边形,平行四边形ABCD中,ABCD, OAEOCF,OAOC 又COFAOE, AOECOF,AECF,又AECF, 四边形AECF是平行四边形,课后巩固,(1)四边形ABCD是平行四边形,ABCD,ABCD.ABECDF. 又BEDF,ABECDF;,4.如下图,ABCD中,点E、F在对角线BD上,且BEDF. (1)求证:ABECDF;(2)求证:四边形AECF是平行四边形,课后巩固,(

4、2)ABEDCF,AEBCFD,AEFCFE,AECF,AECF,四边形AECF是平行四边形,4.如下图,ABCD中,点E、F在对角线BD上,且BEDF. (1)求证:ABECDF;(2)求证:四边形AECF是平行四边形,课后巩固,5.已知:如下图所示,四边形ABCD是平行四边形,E、F是直线BD上的两点,且DEBF. (1)求证:AECF;(2)连接AF、CE,则四边形AFCE是平行四边形吗?说明理由.,(1)四边形ABCD是平行四边形,ADBC,ADBC.ADBCBD.ADECBF.又DEBF,ADECBF.AECF.,课后巩固,(2)四边形AFCE是平行四边形理由如下:ADECBF,AE

5、DCFB.AECF,又AECF,四边形AFCE是平行四边形.,5.已知:如下图所示,四边形ABCD是平行四边形,E、F是直线BD上的两点,且DEBF. (1)求证:AECF;(2)连接AF、CE,则四边形AFCE是平行四边形吗?说明理由.,课后巩固,(1)ACDF,ACDF,四边形ACFD是平行四边形;,(2)由(1)得四边形ACFD是平行四边形, CFAD,CFAD,ADBE,CFBE, 四边形CBEF是平行四边形,BCFE.,6.如下图,点A、D、B、E在一直线上,ADBE,ACDF,ACDF.(1)求证:四边形ACFD是平行四边形;(2)求证:BCFE.,课后巩固,7.如下图,在ABCD

6、中,E、F是对角线BD上的两点,BEDF,点G、H分别在BA和DC的延长线上,且AGCH,连接GE、EH、HF、FG.求证:(1)BEGDFH; (2)四边形GEHF是平行四边形,(1)四边形ABCD是平行四边形,ABCD,ABDC,ABECDF,AGCH,BGDH,又BEDF,BEGDFH;,课后巩固,7.如下图,在ABCD中,E、F是对角线BD上的两点,BEDF,点G、H分别在BA和DC的延长线上,且AGCH,连接GE、EH、HF、FG.求证:(1)BEGDFH; (2)四边形GEHF是平行四边形,(2)BEGDFH,BEGDFH,EGFH,GEFHFB,GEFH,四边形GEHF是平行四边

7、形,课后巩固,8.如下图,已知在平行四边形ABCD中,E,F为边AD, BC上的点,且AECF,连接AF,EC,BE,DF交于M,N,求证:四边形EMFN是平行四边形,四边形ABCD是平行四边形,则ADBC. ADBC,DEBF且DEBF, 四边形BEDF是平行四边形, BEDF,即MEFN. 又AECF且AECF, 四边形AECF是平行四边形,MFEN, 四边形MENF是平行四边形,,能力培优,9.如下图,分别以RtABC的直角边AC及斜边AB向外作等边ACD、等边 ABE.已知BAC30, EFAB,垂足为F,连结DF.(1)求证:ACEF;(2)求证:四边形ADFE是平行四边形,(1)等

8、边ABE,ABE60,ABBE,EFAB,BFEAFE90,BAC30,ACB90,ABC60,ABCABE,ACBBFE90,ABCEFB,ACEF;,能力培优,(2)等边ACD,ADAC,CAD60,BAD90,ADEF,ACEF,ADEF,四边形ADFE是平行四边形,9.如下图,分别以RtABC的直角边AC及斜边AB向外作等边ACD、等边 ABE.已知BAC30, EFAB,垂足为F,连结DF.(1)求证:ACEF;(2)求证:四边形ADFE是平行四边形,能力培优,t,12t,152t,2t,能力培优,(2)当t为何值时,四边形APQB是平行四边形?,(2)根据题意有APt,CQ2t,PD12t,BQ152t.ADBC,当APBQ时,四边形APQB是平行四边形t152t,解得t5.t5 s时四边形APQB是平行四边形;,能力培优,(3)当t为何值时,四边形PDCQ是平行四边形?,(3)由APt cm,CQ2t cm,AD12 cm,BC15 cm,PDADAP12t,如下图,ADBC,当PDQC时,四边形PDCQ是平行四边形即:12t2t,解得t4 s,当t4 s时,四边形PDCQ是平行四边形,感谢聆听,

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