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八年级数学填空题一般
题目
如图,ABC\triangle ABC中,C=90\angle C=90^{\circ},ADAD平分BAC\angle BAC,EEACAC边上的点,连接DEDE,DE=DBDE=DB,下列结论:
DEA+B=180\angle DEA+\angle B=180^{\circ};②ABAC=CEAB-AC=CE;③AC=12(AB+CD)AC=\frac{1}{2}(AB+CD);④SADC=12S四边形ABDES_{\triangle ADC}=\frac{1}{2}S_{四边形ABDE},其中一定正确的结论有______(填写序号即可)(填写序号即可).
知识点:角的运算、三角形内角和定理、全等三角形的性质、全等三角形的判定、等腰三角形的性质、勾股定理、旋转的性质章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

如图,过DDDFABDF\bot ABFF

C=90\because \angle C=90^{\circ}ADAD是角平分线,
DC=DF\therefore DC=DFC=DFB\angle C=\angle DFB
DE=DB\because DE=DB
RtCDE\therefore Rt\triangle CDERtFDB(HL)Rt\triangle FDB\left(HL\right)
B=CED\therefore \angle B=\angle CEDCDE=FDB\angle CDE=\angle FDBCE=BFCE=BF
DEA+DEC=180\because \angle DEA+\angle DEC=180^{\circ}
DEA+B=180\therefore \angle DEA+\angle B=180^{\circ},故①正确;
AD=AD\because AD=ADDC=DFDC=DF
RtCDA\therefore Rt\triangle CDARtFDA(HL)Rt\triangle FDA\left(HL\right)
AC=AF\therefore AC=AF
ABAC=ABAF=BF=CE\therefore AB-AC=AB-AF=BF=CE,故②正确;
AC=AF\because AC=AF
AB+AE=(AF+FB)+(ACCE)=AF+AC=2AC\therefore AB+AE=\left(AF+FB\right)+\left(AC-CE\right)=AF+AC=2AC
AC=12(AB+AE)\therefore AC=\frac{1}{2}(AB+AE)
CDAE\because CD\neq AE
AC12(AB+CD)\therefore AC\neq \frac{1}{2}(AB+CD),故③错误;
RtCDE\because Rt\triangle CDERtFDBRt\triangle FDB
SCDE=SFDB\therefore S_{\triangle CDE}=S_{\triangle FDB}
S四边形ABDE=S四边形ACDF\therefore S_{四边形ABDE}=S_{四边形ACDF}
ACD\because \triangle ACDAFD\triangle AFD
SACD=SADF\therefore S_{\triangle ACD}=S_{\triangle ADF}
SADC=12S四边形ACDF=12S四边形ABDE\therefore S_{\triangle ADC}=\frac{1}{2}S_{四边形ACDF}=\frac{1}{2}S_{四边形ABDE},故④正确;
\therefore一定正确的结论有①②④.
故答案为:①②④.

解析

如图,过DDDFABDF\bot ABFF

C=90\because \angle C=90^{\circ}ADAD是角平分线,
DC=DF\therefore DC=DFC=DFB\angle C=\angle DFB
DE=DB\because DE=DB
RtCDE\therefore Rt\triangle CDERtFDB(HL)Rt\triangle FDB\left(HL\right)
B=CED\therefore \angle B=\angle CEDCDE=FDB\angle CDE=\angle FDBCE=BFCE=BF
DEA+DEC=180\because \angle DEA+\angle DEC=180^{\circ}
DEA+B=180\therefore \angle DEA+\angle B=180^{\circ},故①正确;
AD=AD\because AD=ADDC=DFDC=DF
RtCDA\therefore Rt\triangle CDARtFDA(HL)Rt\triangle FDA\left(HL\right)
AC=AF\therefore AC=AF
ABAC=ABAF=BF=CE\therefore AB-AC=AB-AF=BF=CE,故②正确;
AC=AF\because AC=AF
AB+AE=(AF+FB)+(ACCE)=AF+AC=2AC\therefore AB+AE=\left(AF+FB\right)+\left(AC-CE\right)=AF+AC=2AC
AC=12(AB+AE)\therefore AC=\frac{1}{2}(AB+AE)
CDAE\because CD\neq AE
AC12(AB+CD)\therefore AC\neq \frac{1}{2}(AB+CD),故③错误;
RtCDE\because Rt\triangle CDERtFDBRt\triangle FDB
SCDE=SFDB\therefore S_{\triangle CDE}=S_{\triangle FDB}
S四边形ABDE=S四边形ACDF\therefore S_{四边形ABDE}=S_{四边形ACDF}
ACD\because \triangle ACDAFD\triangle AFD
SACD=SADF\therefore S_{\triangle ACD}=S_{\triangle ADF}
SADC=12S四边形ACDF=12S四边形ABDE\therefore S_{\triangle ADC}=\frac{1}{2}S_{四边形ACDF}=\frac{1}{2}S_{四边形ABDE},故④正确;
\therefore一定正确的结论有①②④.
故答案为:①②④.

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