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八年级数学填空题一般
题目
新知学习:若一条线段把一个平面图形分成面积相等的两部分,我们把这条线段叫做该平面图形的二分线.
解决问题:
(1)(1)①三角形的中线、高线、角平分线中,一定是三角形的二分线的是______;
②如图11,已知ABC\triangle ABC中,ADADBCBC边上的中线,点EE,FF分别在ABAB,DCDC上,连接EFEF,与ADAD交于点GG.若SAEG=SDGFS_{\triangle AEG}=S_{\triangle DGF},则EFEF______(填"是"或"不是")ABC)\triangle ABC的一条二分线.
(2)(2)如图22,四边形ABCDABCD中,CDCD平行于ABAB,点GGADAD的中点,射线CGCG交射线BABA于点EE,取EBEB的中点FF,连接CFCF.求证:CFCF是四边形ABCDABCD的二分线.
(3)(3)如图33,在ABC\triangle ABC中,AB=CB=CE=7AB=CB=CE=7,A=C\angle A=\angle C,CBE=CEB\angle CBE=\angle CEB,DD,EE分别是线段BCBC,ACAC上的点,且BED=A\angle BED=\angle A,EFEF是四边形ABDEABDE的一条二分线,求DFDF的长.
知识点:三角形章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

(1)\left(1\right)\because三角形的中线把三角形分成面积相等的两部分;
\therefore三角形的中线是三角形的二分线,
故答案为三角形的中线
AD\because ADBCBC边上的中线
SABD=SACD=12SABC\therefore S_{\triangle ABD}=S_{\triangle ACD}=\frac{1}{2}S_{\triangle ABC}
SAEG=SDGF\because S_{\triangle AEG}=S_{\triangle DGF}
S四边形BDGE+SAEG=S四边形BDGE+SDGF\therefore S_{四边形BDGE}+S_{\triangle AEG}=S_{四边形BDGE}+S_{\triangle DGF}
SBEF=SABD=12SABC\therefore S_{\triangle BEF}=S_{\triangle ABD}=\frac{1}{2}S_{\triangle ABC}
EF\therefore EFABC\triangle ABC的一条二分线
故答案为:是
(2)EB(2)\because EB的中点FF
SCBF=SCEF\therefore S_{\triangle CBF}=S_{\triangle CEF}
AB\because ABDCDC
E=DCG\therefore \angle E=\angle DCG
G\because GADAD的中点,
DG=AG\therefore DG=AG
CDG\triangle CDGEAG\triangle EAG中,
{E=DCGEGA=CGDAG=DG\left\{\begin{array}{l}{∠E=∠DCG}\\{∠EGA=∠CGD}\\{AG=DG}\end{array}\right.
CDG\therefore \triangle CDGEAG(AAS)\triangle EAG\left(AAS\right)
SAEG=SDCG\therefore S_{\triangle AEG}=S_{\triangle DCG}
S四边形AFCD=SCEF\therefore S_{四边形AFCD}=S_{\triangle CEF}
S四边形AFCD=SCBF\therefore S_{四边形AFCD}=S_{\triangle CBF}
CF\therefore CF是四边形ABCDABCD的二分线.
(3)(3)如图,延长CBCB使BH=CDBH=CD,连接EHEH

AB=CB=CE=7AB=CB=CE=7A=C\angle A=\angle CCBE=CEB\angle CBE=\angle CEBDDEE分别是线段BCBCACAC上的点,且BED=A\angle BED=\angle A
BC=7\because BC=7
BD+CD=7\therefore BD+CD=7
BD+BH=7=HD\therefore BD+BH=7=HD
BED=A\because \angle BED=\angle ABED+DEC=A+ABE\angle BED+\angle DEC=\angle A+\angle ABE
ABE=CED\therefore \angle ABE=\angle CED,且AB=CE=7AB=CE=7A=C\angle A=\angle C
ABE\therefore \triangle ABECED(ASA)\triangle CED\left(ASA\right)
AE=CD\therefore AE=CDBE=DEBE=DEAEB=EDC\angle AEB=\angle EDCSABE=SEDCS_{\triangle ABE}=S_{\triangle EDC}
AE=BH\therefore AE=BH
CBE=CEB\because \angle CBE=\angle CEB
AEB=EBH\therefore \angle AEB=\angle EBH
EBH=EDC\therefore \angle EBH=\angle EDC,且BE=DEBE=DEBH=CDBH=CD
BEH\therefore \triangle BEHDEC(SAS)\triangle DEC\left(SAS\right)
SBEH=SDEC\therefore S_{\triangle BEH}=S_{\triangle DEC}
SBEH=SDEC=SABE\therefore S_{\triangle BEH}=S_{\triangle DEC}=S_{\triangle ABE}
SHED=S四边形ABDE\therefore S_{\triangle HED}=S_{四边形ABDE}
EF\because EF是四边形ABDEABDE的一条二分线,
SDEF=12S四边形ABDE=12SHED\therefore S_{\triangle DEF}=\frac{1}{2}S_{四边形ABDE}=\frac{1}{2}S_{\triangle HED}
DF=12DH=72\therefore DF=\frac{1}{2}DH=\frac{7}{2}

解析

(1)\left(1\right)\because三角形的中线把三角形分成面积相等的两部分;
\therefore三角形的中线是三角形的二分线,
故答案为三角形的中线
AD\because ADBCBC边上的中线
SABD=SACD=12SABC\therefore S_{\triangle ABD}=S_{\triangle ACD}=\frac{1}{2}S_{\triangle ABC}
SAEG=SDGF\because S_{\triangle AEG}=S_{\triangle DGF}
S四边形BDGE+SAEG=S四边形BDGE+SDGF\therefore S_{四边形BDGE}+S_{\triangle AEG}=S_{四边形BDGE}+S_{\triangle DGF}
SBEF=SABD=12SABC\therefore S_{\triangle BEF}=S_{\triangle ABD}=\frac{1}{2}S_{\triangle ABC}
EF\therefore EFABC\triangle ABC的一条二分线
故答案为:是
(2)EB(2)\because EB的中点FF
SCBF=SCEF\therefore S_{\triangle CBF}=S_{\triangle CEF}
AB\because ABDCDC
E=DCG\therefore \angle E=\angle DCG
G\because GADAD的中点,
DG=AG\therefore DG=AG
CDG\triangle CDGEAG\triangle EAG中,
{E=DCGEGA=CGDAG=DG\left\{\begin{array}{l}{∠E=∠DCG}\\{∠EGA=∠CGD}\\{AG=DG}\end{array}\right.
CDG\therefore \triangle CDGEAG(AAS)\triangle EAG\left(AAS\right)
SAEG=SDCG\therefore S_{\triangle AEG}=S_{\triangle DCG}
S四边形AFCD=SCEF\therefore S_{四边形AFCD}=S_{\triangle CEF}
S四边形AFCD=SCBF\therefore S_{四边形AFCD}=S_{\triangle CBF}
CF\therefore CF是四边形ABCDABCD的二分线.
(3)(3)如图,延长CBCB使BH=CDBH=CD,连接EHEH

AB=CB=CE=7AB=CB=CE=7A=C\angle A=\angle CCBE=CEB\angle CBE=\angle CEBDDEE分别是线段BCBCACAC上的点,且BED=A\angle BED=\angle A
BC=7\because BC=7
BD+CD=7\therefore BD+CD=7
BD+BH=7=HD\therefore BD+BH=7=HD
BED=A\because \angle BED=\angle ABED+DEC=A+ABE\angle BED+\angle DEC=\angle A+\angle ABE
ABE=CED\therefore \angle ABE=\angle CED,且AB=CE=7AB=CE=7A=C\angle A=\angle C
ABE\therefore \triangle ABECED(ASA)\triangle CED\left(ASA\right)
AE=CD\therefore AE=CDBE=DEBE=DEAEB=EDC\angle AEB=\angle EDCSABE=SEDCS_{\triangle ABE}=S_{\triangle EDC}
AE=BH\therefore AE=BH
CBE=CEB\because \angle CBE=\angle CEB
AEB=EBH\therefore \angle AEB=\angle EBH
EBH=EDC\therefore \angle EBH=\angle EDC,且BE=DEBE=DEBH=CDBH=CD
BEH\therefore \triangle BEHDEC(SAS)\triangle DEC\left(SAS\right)
SBEH=SDEC\therefore S_{\triangle BEH}=S_{\triangle DEC}
SBEH=SDEC=SABE\therefore S_{\triangle BEH}=S_{\triangle DEC}=S_{\triangle ABE}
SHED=S四边形ABDE\therefore S_{\triangle HED}=S_{四边形ABDE}
EF\because EF是四边形ABDEABDE的一条二分线,
SDEF=12S四边形ABDE=12SHED\therefore S_{\triangle DEF}=\frac{1}{2}S_{四边形ABDE}=\frac{1}{2}S_{\triangle HED}
DF=12DH=72\therefore DF=\frac{1}{2}DH=\frac{7}{2}

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