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八年级数学填空题一般
题目
(1)(1)如图1,AE1,AEBCBC,AEAE平分DAC\angle DAC,则ABC\triangle ABC的形状是______三角形.
(2)(2)如图22,BCBC平分ABD,AC\angle ABD,ACBDBD,AC=3AC=3,则AB=______.AB=\_\_\_\_\_\_.
(3)(3)如图33,有ABC\triangle ABC中,BEBE是角平分线,DE,DEBCBCABAB于点DD.若DE=7DE=7,AD=5AD=5,则AB=______.AB=\_\_\_\_\_\_.
(4)(4)如图44,在ABC\triangle ABC中,ABC\angle ABCACB\angle ACB的平分线交于点FF,过点FFDEDEBCBC,分别交ABAB,ACAC于点DD,EE.若AB=12AB=12,AC=18AC=18,BC=24BC=24,则ADE\triangle ADE的周长为______.
(5)(5)如图55,在ABC\triangle ABC中,BC=5cmBC=5cm,BPBP,CPCP分别是ABC\angle ABCACB\angle ACB的平分线,且PDPDAB,PEAB,PEACAC,则PDE\triangle PDE的周长是______.
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)AE\left(1\right)\because AEBCBC
DAE=B\therefore \angle DAE=\angle BEAC=C\angle EAC=\angle C
AE\because AE平分DAC\angle DAC
DAE=EAC\therefore \angle DAE=\angle EAC
B=C\therefore \angle B=\angle C
AB=AC\therefore AB=AC
ABC\therefore \triangle ABC是等腰三角形,
故答案为:等腰;
(2)BC(2)\because BC平分ABD\angle ABD
ABC=CBD\therefore \angle ABC=\angle CBD.
AC\because ACBDBD
C=CBD\therefore \angle C=\angle CBD
C=ABC\therefore \angle C=\angle ABC
AB=AC\therefore AB=AC.
AC=3\because AC=3
AB=3\therefore AB=3.
故答案为:33.
(3)BE(3)\because BE平分ABC\angle ABC
ABE=EBC\therefore \angle ABE=\angle EBC.
DE\because DEBCBC
DEB=DBE\therefore \angle DEB=\angle DBE
DEB=ABE\therefore \angle DEB=\angle ABE
DB=DE\therefore DB=DE.
DE=7\because DE=7AD=5AD=5
AB=AD+DB=AD+DE=12\therefore AB=AD+DB=AD+DE=12.
故答案为:1212.
(4)BF(4)\because BF平分ABC\angle ABC
ABF=FBC\therefore \angle ABF=\angle FBC
DE\because DEBCBC
FBC=DFB\therefore \angle FBC=\angle DFB
ABF=DFB\therefore \angle ABF=\angle DFB
DB=DF\therefore DB=DF.
同理可得,CE=EFCE=EF.
AB=12\because AB=12AC=18AC=18
CADE=AD+DE+AE\therefore C_{\triangle ADE}=AD+DE+AE
=AD+DF+EF+AE=AD+DF+EF+AE
=AD+DB+CE+AE=AD+DB+CE+AE
=AB+AC=AB+AC
=12+18=12+18
=30=30.
故答案为:3030.
(5)BP(5)\because BP平分ABC\angle ABC
ABP=PBC\therefore \angle ABP=\angle PBC
PD\because PDABAB
ABP=BPD\therefore \angle ABP=\angle BPD
PBC=BPD\therefore \angle PBC=\angle BPD
DP=DB\therefore DP=DB.
同理可得,EP=ECEP=EC.
BC=5cm\because BC=5cm
CPDE=PD+DE+EP=BD+DE+EC=BC=5(cm)\therefore C_{\triangle PDE}=PD+DE+EP=BD+DE+EC=BC=5\left(cm\right).
故答案为:5cm5cm.

解析

(1)AE\left(1\right)\because AEBCBC
DAE=B\therefore \angle DAE=\angle BEAC=C\angle EAC=\angle C
AE\because AE平分DAC\angle DAC
DAE=EAC\therefore \angle DAE=\angle EAC
B=C\therefore \angle B=\angle C
AB=AC\therefore AB=AC
ABC\therefore \triangle ABC是等腰三角形,
故答案为:等腰;
(2)BC(2)\because BC平分ABD\angle ABD
ABC=CBD\therefore \angle ABC=\angle CBD.
AC\because ACBDBD
C=CBD\therefore \angle C=\angle CBD
C=ABC\therefore \angle C=\angle ABC
AB=AC\therefore AB=AC.
AC=3\because AC=3
AB=3\therefore AB=3.
故答案为:33.
(3)BE(3)\because BE平分ABC\angle ABC
ABE=EBC\therefore \angle ABE=\angle EBC.
DE\because DEBCBC
DEB=DBE\therefore \angle DEB=\angle DBE
DEB=ABE\therefore \angle DEB=\angle ABE
DB=DE\therefore DB=DE.
DE=7\because DE=7AD=5AD=5
AB=AD+DB=AD+DE=12\therefore AB=AD+DB=AD+DE=12.
故答案为:1212.
(4)BF(4)\because BF平分ABC\angle ABC
ABF=FBC\therefore \angle ABF=\angle FBC
DE\because DEBCBC
FBC=DFB\therefore \angle FBC=\angle DFB
ABF=DFB\therefore \angle ABF=\angle DFB
DB=DF\therefore DB=DF.
同理可得,CE=EFCE=EF.
AB=12\because AB=12AC=18AC=18
CADE=AD+DE+AE\therefore C_{\triangle ADE}=AD+DE+AE
=AD+DF+EF+AE=AD+DF+EF+AE
=AD+DB+CE+AE=AD+DB+CE+AE
=AB+AC=AB+AC
=12+18=12+18
=30=30.
故答案为:3030.
(5)BP(5)\because BP平分ABC\angle ABC
ABP=PBC\therefore \angle ABP=\angle PBC
PD\because PDABAB
ABP=BPD\therefore \angle ABP=\angle BPD
PBC=BPD\therefore \angle PBC=\angle BPD
DP=DB\therefore DP=DB.
同理可得,EP=ECEP=EC.
BC=5cm\because BC=5cm
CPDE=PD+DE+EP=BD+DE+EC=BC=5(cm)\therefore C_{\triangle PDE}=PD+DE+EP=BD+DE+EC=BC=5\left(cm\right).
故答案为:5cm5cm.

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