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九年级数学解答题一般
题目
如图,已知ADE\triangle ADE的顶点EEABC\triangle ABC的边BCBC上,DEDEABAB相交于点FF,DAE=CAB\angle DAE=\angle CAB,AE2=AFABAE^{2}=AF\cdot AB.
(1)(1)求证:FEA=B\angle FEA=\angle B
(2)(2)求证:DFAD=CEAC\frac{DF}{AD}=\frac{CE}{AC}.
知识点:等腰三角形的性质、勾股定理、直角三角形的性质、圆周角定理I、切线的性质章节:第22章 圆(下) / 一 直线和圆 / 22.2 圆的切线

答案与解析

答案

证明:(1)AFAB=AE2\left(1\right)\because AF\cdot AB=AE^{2}

AEAF=ABAE\therefore \frac{AE}{AF}=\frac{AB}{AE}

EAB=FAE\because \angle EAB=\angle FAE

EAF\therefore \triangle EAFBAE\triangle BAE

FEA=B\therefore \angle FEA=\angle B

(2)FEA=B(2)\because \angle FEA=\angle BDAE=CAB\angle DAE=\angle CAB

DAF=EAC,DAE\therefore \angle DAF=\angle EAC,\triangle DAECAB\triangle CAB

D=C\therefore \angle D=\angle CDEBC=ADAC\frac{DE}{BC}=\frac{AD}{AC}

DAF=EAC\because \angle DAF=\angle EAC

DAF\therefore \triangle DAFCAE\triangle CAE

DAAC=DFCE\therefore \frac{DA}{AC}=\frac{DF}{CE}

DFAD=CEAC\therefore \frac{DF}{AD}=\frac{CE}{AC}.

解析

证明:(1)AFAB=AE2\left(1\right)\because AF\cdot AB=AE^{2}

AEAF=ABAE\therefore \frac{AE}{AF}=\frac{AB}{AE}

EAB=FAE\because \angle EAB=\angle FAE

EAF\therefore \triangle EAFBAE\triangle BAE

FEA=B\therefore \angle FEA=\angle B

(2)FEA=B(2)\because \angle FEA=\angle BDAE=CAB\angle DAE=\angle CAB

DAF=EAC,DAE\therefore \angle DAF=\angle EAC,\triangle DAECAB\triangle CAB

D=C\therefore \angle D=\angle CDEBC=ADAC\frac{DE}{BC}=\frac{AD}{AC}

DAF=EAC\because \angle DAF=\angle EAC

DAF\therefore \triangle DAFCAE\triangle CAE

DAAC=DFCE\therefore \frac{DA}{AC}=\frac{DF}{CE}

DFAD=CEAC\therefore \frac{DF}{AD}=\frac{CE}{AC}.

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