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八年级数学填空题一般
题目
(1)(1)如图11,ABC\triangle ABC的三条边相等,三个内角也相等,点DDEEFF分别在边ABABBCBCCACA上,且BD=CE=AFBD=CE=AF.请写出图中一对全等三角形______,其全等的理由是______;

(2)(2)如图22,ABC\triangle ABC中,AB=ACAB=AC,点DDEEFF分别在边ABABBCBCACAC上,且BD=CEBD=CE,DEF=B\angle DEF=\angle B,请判断DEF\triangle DEF的形状,并说明理由;
(3)(3)如图33,ABC\triangle ABC中,AB=AC=8AB=AC=8,点DDBABA的延长线上,点EE在边BCBC上,且AD=CE=2AD=CE=2,DEF=B\angle DEF=\angle B.延长BCBC至点MM,使得CM=CACM=CA,过点MMACAC的平行线MFMF,与边EFEF交于点FF.若MF=4MF=4,请你求出线段BMBM的长度.
知识点:线段垂直平分线的性质、全等三角形的判定、等腰三角形的性质、等边三角形的性质章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

(1)由题意得:AB=AC=BCAB=AC=BCA=B=C\angle A=\angle B=\angle C
BD=CE=AF\because BD=CE=AF
AD=BE\therefore AD=BE
ADF\triangle ADFBED\triangle BED中,
{AF=BDA=BAD=BE\left\{\begin{array}{c}AF=BD\\∠A=∠B\\ AD=BE\end{array}\right.
,ADF,\therefore \triangle ADFBED(SAS)\triangle BED\left(SAS\right)
故答案为:ADF\triangle ADFBED(答案不唯一)\triangle BED(答案不唯一)SASSAS
(2)DEF(2)\triangle DEF为等腰三角形,
理由如下:AB=AC\because AB=AC
B=C\therefore \angle B=\angle C
DEC=B+BDE=DEF+CEF\because \angle DEC=\angle B+\angle BDE=\angle DEF+\angle CEFDEF=B\angle DEF=\angle B
BDE=CEF\therefore \angle BDE=\angle CEF
BDE\triangle BDECEF\triangle CEF中,
{BDE=CEFBD=CEB=C\left\{\begin{array}{c}∠BDE=∠CEF\\ BD=CE\\∠B=∠C\end{array}\right.
,BDE,\therefore \triangle BDECEF(ASA)\triangle CEF\left(ASA\right)
DE=EF\therefore DE=EF
DEF\therefore \triangle DEF为等腰三角形;
(3)AB=AC(3)\because AB=AC
B=ACB\therefore \angle B=\angle ACB
AC\because ACFMFM
M=ACB\therefore \angle M=\angle ACB
B=M\therefore \angle B=\angle M
AB=AC\because AB=ACCM=CACM=CA
AB=CM\therefore AB=CM
AD=CE\because AD=CE
AB+AD=CM+CE\therefore AB+AD=CM+CE,即BD=MEBD=ME
由(2)可知:DEF=B\angle DEF=\angle B时,D=MEF\angle D=\angle MEF
DBE\triangle DBEEMF\triangle EMF中,
{B=MBD=EMD=MEF\left\{\begin{array}{c}∠B=∠M\\ BD=EM\\∠D=∠MEF\end{array}\right.
,DBE,\therefore \triangle DBEEMF(ASA)\triangle EMF\left(ASA\right)
BE=MF=4\therefore BE=MF=4EM=BD=AB+AD=10EM=BD=AB+AD=10
BD=BE+EM=4+10=14\therefore BD=BE+EM=4+10=14.

解析

(1)由题意得:AB=AC=BCAB=AC=BCA=B=C\angle A=\angle B=\angle C
BD=CE=AF\because BD=CE=AF
AD=BE\therefore AD=BE
ADF\triangle ADFBED\triangle BED中,
{AF=BDA=BAD=BE\left\{\begin{array}{c}AF=BD\\∠A=∠B\\ AD=BE\end{array}\right.
,ADF,\therefore \triangle ADFBED(SAS)\triangle BED\left(SAS\right)
故答案为:ADF\triangle ADFBED(答案不唯一)\triangle BED(答案不唯一)SASSAS
(2)DEF(2)\triangle DEF为等腰三角形,
理由如下:AB=AC\because AB=AC
B=C\therefore \angle B=\angle C
DEC=B+BDE=DEF+CEF\because \angle DEC=\angle B+\angle BDE=\angle DEF+\angle CEFDEF=B\angle DEF=\angle B
BDE=CEF\therefore \angle BDE=\angle CEF
BDE\triangle BDECEF\triangle CEF中,
{BDE=CEFBD=CEB=C\left\{\begin{array}{c}∠BDE=∠CEF\\ BD=CE\\∠B=∠C\end{array}\right.
,BDE,\therefore \triangle BDECEF(ASA)\triangle CEF\left(ASA\right)
DE=EF\therefore DE=EF
DEF\therefore \triangle DEF为等腰三角形;
(3)AB=AC(3)\because AB=AC
B=ACB\therefore \angle B=\angle ACB
AC\because ACFMFM
M=ACB\therefore \angle M=\angle ACB
B=M\therefore \angle B=\angle M
AB=AC\because AB=ACCM=CACM=CA
AB=CM\therefore AB=CM
AD=CE\because AD=CE
AB+AD=CM+CE\therefore AB+AD=CM+CE,即BD=MEBD=ME
由(2)可知:DEF=B\angle DEF=\angle B时,D=MEF\angle D=\angle MEF
DBE\triangle DBEEMF\triangle EMF中,
{B=MBD=EMD=MEF\left\{\begin{array}{c}∠B=∠M\\ BD=EM\\∠D=∠MEF\end{array}\right.
,DBE,\therefore \triangle DBEEMF(ASA)\triangle EMF\left(ASA\right)
BE=MF=4\therefore BE=MF=4EM=BD=AB+AD=10EM=BD=AB+AD=10
BD=BE+EM=4+10=14\therefore BD=BE+EM=4+10=14.

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