
2025年广西九年级中考数学一轮复习课件:第38讲几何类.pptx
32页第38讲几何类,1,.,了解初中几何常见模型,,,如手拉手全等相似模型,、,对角互补,模型,、,倍长中线模型和,123,模型等,.,2,.,通过图形建模实现知识多元化认知,,,顺利找到解题的方法,.,3,.,通过观察与联想,,,构造出图形模型,,,把复杂的问题转化为简,单的问题求解,.,1,.,【手拉手全等模型】,如图,,,C,为线,段,A,E,上一动点,(,不与,点,A,,,E,重合,),,,在,A,E,同侧分别作等边三角,形,AB,C,和等边三角,形,CD,E,,,A,D,与,B,E,交于,点,O,,,A,D,与,B,C,交于,点,P,,,B,E,与,C,D,交于,点,Q,,,连,接,P,Q,.,求证,:,(,1,),BE,C,AD,C,;,证明,:,AB,C,和,DC,E,为等边三角形,,,B,C,A,C,,,C,D,C,E,,,BC,A,DC,E,60,.,AC,D,BC,E,.,在,BE,C,和,AD,C,中,,,,,BE,C,AD,C,(,SAS,),.,(,2,),PQ,C,是等边三角形,.,证明,:,AD,C,BE,C,,,AD,C,BE,C,.,AC,B,DC,E,60,,,BC,D,60,.,在,DP,C,和,EQ,C,中,,,DP,C,EQ,C,(,ASA,),.,C,P,C,Q,.,QC,P,60,,,PQ,C,是等边三角形,.,2,.,【手拉手相似模型】,如图,,,在,AB,C,和,AD,E,中,,,,,BA,D,20,,,求,CA,E,的度数,.,解,:,在,AB,C,和,AD,E,中,,,,,AB,C,AD,E,.,BA,C,DA,E,.,BA,C,DA,C,DA,E,DA,C,.,CA,E,BA,D,20,.,3,.,【倍长中线模型】,如图,,,在,AB,C,中,,,若,A,B,6,,,A,C,4,,,求,B,C,边上的中,线,A,D,的取值范围,.,解,:,如图,,,延,长,A,D,到,点,E,,,使,D,E,A,D,,,连,接,B,E,.,A,D,为,B,C,边上中线,,,C,D,B,D,.,在,DA,C,和,DE,B,中,,,,,DA,C,DE,B,(,SAS,),.,A,C,E,B,4,.,A,B,B,E,A,E,A,B,B,E,,,A,B,6,,,2,A,E,10,.,1,A,D,5,.,B,C,边上的中,线,A,D,的取值范围是,1,A,D,5,.,4,.,【对角互补模型】,如图,,,在四边,形,ABC,D,中,,,A,B,A,D,,,A,C,5,,,DA,B,DC,B,90,,,则四边,形,ABC,D,的面积为,(,B,),A,.,15,B,.,12,.,5,C,.,14,.,5,D,.,17,B,5,.,【,123,模型】,如图,,,用,6,个边长为,1,的小正方形构造的网格图,,,角,,,的顶点均在格点上,,,则,.,45,典型考题,将,AB,C,绕着,点,A,逆时针旋转得到,AD,E,.,(,1,),如图,,,当,点,D,恰好落,在,B,C,上时,,,连,接,C,E,.,当,B,70,,,AC,B,50,时,,,求证,:,A,C,D,E,.,证明,:,如图,,,设,A,C,,,E,D,相交于,点,F,.,由旋转,,,可得,AB,C,AD,E,.,B,AD,E,70,,,A,B,A,D,.,B,AD,B,70,.,FD,C,180,AD,B,AD,E,40,.,又,AC,B,50,,,CF,D,180,FD,C,AC,B,180,40,50,90,.,A,C,D,E,.,解,:,当,AB,C,中,,,B,C,A,C,时,,,四边,形,ABC,E,是平行四边形,.,理由如下,:,如图,,,BA,C,DA,E,,,BA,C,2,DA,E,2,,,即,1,3,.,当,AB,C,满足什么条件时,,,四边,形,ABC,E,是平行四边形,?,说明,理由,.,B,C,A,C,,,B,BA,C,.,5,180,2,B,.,同理,,,可得,1,180,2,B,.,5,1,.,5,3,.,A,E,B,C,.,又,B,C,A,C,,,A,C,A,E,,,B,C,A,E,.,四边,形,ABC,E,是平行四边形,.,(,2,),如图,,,当旋转角为,60,时,,,D,E,交,B,C,于,点,P,,,连,接,A,P,.,当,A,C,6,,,C,45,时,,,求,P,E,的长,.,解,:,如图,,,过,点,A,作,A,M,B,C,于,点,M,,,A,N,D,E,于,点,N,.,设,DE,与,A,C,交于,点,H,.,AN,E,AM,C,90,.,C,E,45,,,A,C,A,E,,,AM,C,AN,E,(,AAS,),.,A,M,A,N,.,P,A,平分,MP,N,,,即,AP,N,MP,N,.,C,E,,,AH,E,PH,C,,,旋转角为,60,,,HP,C,HA,E,60,.,BP,H,180,HP,C,180,60,120,.,AP,N,MP,N,120,60,.,在等腰直角三角,形,AN,E,中,,,A,E,A,C,6,,,N,E,A,N,3,.,在,Rt,AP,N,中,,,AP,N,60,,,PA,N,30,.,P,N,.,P,E,P,N,N,E,3,.,变式训练,课题学习,:,三角形旋转问题中的,“,转化思想,”,(,1,),如图,,,在等腰,AB,C,中,,,A,C,A,B,,,CA,B,90,,,点,D,在,AB,C,内部,,,连,接,A,D,,,将,A,D,绕,点,A,顺时针旋转,90,得,到,A,E,,,连,接,D,E,,,C,D,,,B,E,.,请写,出,B,E,和,C,D,的数量关系,:,,,位置关系,:,,,并证明,.,B,E,C,D,B,E,C,D,证明,:,A,C,A,B,,,CA,B,90,,,AB,C,AC,B,45,.,线,段,A,D,绕,点,A,逆时针旋转,90,得,到,A,E,,,A,D,A,E,,,DA,E,90,.,CA,D,DA,B,CA,B,90,,,DA,B,BA,E,DA,E,90,,,CA,D,BA,E,.,CA,D,BA,E,(,SAS,),.,C,D,B,E,,,AC,D,AB,E,.,延,长,C,D,交,B,E,于,点,F,,,如图,所示,.,FC,B,AB,C,AB,E,FC,B,AB,C,AC,D,AB,C,AC,B,90,,,CF,B,180,(,FC,B,AB,C,AB,E,),180,90,90,.,B,E,C,D,.,(,2,),如图,,,在等腰,AB,C,中,,,A,C,B,C,4,,,AC,B,90,,,A,D,2,,,将,A,D,绕,点,A,顺时针旋转,90,得,到,A,E,,,连,接,D,E,,,B,D,,,B,E,,,取,B,D,的中,点,M,,,连,接,C,M,.,当,点,D,在,AB,C,内部,,,猜想并证,明,B,E,与,C,M,的数量关系和位,置关系,;,解,:,B,E,2,C,M,,,B,E,C,M,.,证明如下,:,如图,,,取,A,B,的中点,为,F,,,B,E,的中点,为,P,,,连,接,F,C,,,F,P,,,F,M,,,延,长,C,M,交,B,E,于,点,O,.,又,B,D,的中点,为,M,,,F,P,是,AB,E,的中位线,,,F,M,是,AB,D,的中位线,.,F,P,A,E,,,F,P,A,E,,,F,M,A,D,,,F,M,A,D,.,A,D,绕,点,A,顺时针旋转,90,得,到,A,E,,,A,D,A,E,,,A,D,A,E,.,F,P,F,M,,,F,P,F,M,.,PF,M,90,.,AB,C,是等腰三角形,,,AC,B,90,,,A,B,的中点,为,F,,,C,F,A,B,,,C,F,B,F,.,CF,B,90,.,CF,M,CF,B,BF,M,90,BF,M,,,BF,P,PF,M,BF,M,90,BF,M,,,CF,M,BF,P,.,又,C,F,B,F,,,F,M,F,P,,,CF,M,BF,P,(,SAS,),.,C,M,B,P,,,FC,M,FB,P,.,B,E,的中点,为,P,,,B,E,2,B,P,2,C,M,.,FB,P,CB,A,MC,B,FC,M,CB,A,MC,B,FC,B,CB,A,180,CF,B,180,90,90,,,CO,B,180,(,FB,P,CB,A,MC,B,),180,90,90,.,B,E,C,M,.,当,B,,,M,,,E,三点共线时,(,M,,,E,在,A,B,的下方,),,,请直接写,出,C,M,的长度,.,C,M,.,图形建模就是指建立几何图形模型的整个过程,,,对真实原型,进行提炼、抽象、简单化,,,以及确立、检验、解释、应用、向外,拓展的过程,.,利用观察与联想等思想,,,准确恰当构造出一个或者多个同源,问题相关的辅助条件或问题,.,建立图形模型,,,把复杂的问题化为,简单的问题求解,.,1,.,(,2024,广州,),如图,,,在,AB,C,中,,,A,90,,,A,B,A,C,6,,,D,为,边,B,C,的中点,,,点,E,,,F,分别在,边,A,B,,,A,C,上,,,A,E,C,F,,,则四边,形,AED,F,的面积为,(,C,),A,.,18,B,.,9,C,.,9,D,.,6,C,2,.,(,2024,佛山顺德区期末,),在,AB,C,中,,,C,90,,,点,M,是线,段,B,C,上的一点,,,连,接,A,M,.,(,1,),如图,,,A,C,B,C,,,A,M,是,AB,C,的角平分线,,,M,E,A,B,于,点,E,.,当,C,M,4,时,,,求,A,B,的长,;,解,:,设,A,C,B,C,x,.,A,M,是,AB,C,的角平分线,,,M,E,A,B,,,C,90,,,C,M,M,E,4,,,B,M,x,4,.,A,C,B,C,,,B,BA,C,45,.,BE,M,为等腰直角三角形,.,即,B,M,M,E,.,x,4,4,.,解,得,x,4,4,.,则,A,B,x,8,4,.,C,F,M,E,且,C,F,M,E,.,理由如下,:,如图,.,AB,C,为等腰直角三角形,,,C,O,为中线,,,C,O,A,B,.,又,M,E,A,B,,,M,E,C,O,,,即,C,F,M,E,.,EM,A,MF,C,.,由,知,,,E,M,C,M,,,A,M,A,M,,,Rt,AM,E,Rt,AM,C,(,HL,),.,EM,A,CM,A,MF,C,.,F,C,C,M,E,M,.,综上所述,,,C,F,M,E,且,C,F,M,E,.,若,AB,C,的中,线,C,O,交,A,M,于,点,F,,,判,断,C,F,与,M,E,的关系,,,并,说明理由,;,(,2,),如图,,,若,B,M,A,C,,,点,N,是,A,C,上的一点,,,且,A,N,C,M,,,连,接,B,N,交,A,M,于,点,P,,,求,BP,M,的度数,.,解,:,如图,,,过,点,M,作,M,E,A,N,,,使,M,E,A,N,,,连,接,N,E,,,B,E,.,四边,形,AME,N,为平行四边形,.,N,E,A,M,,,M,E,B,C,,,1,2,.,A,N,M,C,,,M,E,C,M,.,在,BE,M,和,AM,C,中,,,BE,M,AM,C,(,SAS,),.,B,E,A,M,N,E,,,3,4,.,1,3,90,,,1,2,,,2,4,90,,,即,BE,N,90,.,BE,N,为等腰直角三角形,,,BN,E,45,.,A,M,N,E,,,BP,M,BN,E,45,.,。
