《2025年新编基础训练九年级数学上册北师大版》

6. 如图,四边形$ABCD$为平行四边形,$E为边AD$上一点,连接$AC$,$BE$,它们相交于点$F$,且$\angle ACB= \angle ABE$。
(1)求证：$AE^{2}= EF\cdot BE$；
(2)若$AE= 2$,$EF= 1$,$CF= 4$,求$AB$的长。

答案：
(1)证明:$\because$四边形$ABCD$为平行四边形,
$\therefore AD// BC$,$\therefore \angle DAC=\angle ACB$.
$\because \angle ACB=\angle ABE$,$\therefore \angle DAC=\angle ABE$.
又$\because \angle AEF=\angle BEA$,$\therefore \triangle EAF \backsim \triangle EBA$.
$\therefore EA:EB=EF:EA$.
$\therefore AE^{2}=EF\cdot BE$.
(2)解:$\because AE^{2}=EF\cdot BE$,$\therefore BE=\frac{2^{2}}{1}=4$.
$\therefore BF=BE-EF=4-1=3$.
$\because AE// BC$,$\therefore \frac{AF}{FC}=\frac{EF}{BF}$,即$\frac{AF}{4}=\frac{1}{3}$.
解得$AF=\frac{4}{3}$.
$\because \triangle EAF \backsim \triangle EBA$,
$\therefore \frac{AF}{AB}=\frac{EF}{AE}$,即$\frac{\frac{4}{3}}{AB}=\frac{1}{2}$,$\therefore AB=\frac{8}{3}$.

7. 如图,在$\triangle ABC$中,$BD是AC$边上的高,点$E在边AB$上,连接$CE交BD于点O$,且$AD\cdot OC= AB\cdot OD$。求证：
(1)$CE\perp AB$；
(2)$\triangle DAE\backsim\triangle BAC$。

答案：
(1)证明:$\because AD\cdot OC=AB\cdot OD$,
$\therefore \frac{AD}{OD}=\frac{AB}{OC}$.
$\because BD$是$AC$边上的高,
$\therefore \angle BDC=\angle BDA=90^{\circ}$.
$\therefore \triangle ADB \backsim \triangle ODC$.
$\therefore \angle ABD=\angle OCD$.
又$\because \angle EOB=\angle DOC$,$\angle DOC+\angle OCD+\angle ODC=180^{\circ}$,$\angle EOB+\angle ABD+\angle OEB=180^{\circ}$.
$\therefore \angle OEB=\angle ODC=90^{\circ}$.
$\therefore CE\perp AB$.
(2)证明:在$\triangle ADB$和$\triangle AEC$中,
$\because \angle BAD=\angle CAE$,$\angle ABD=\angle OCD$,
$\therefore \triangle ADB \backsim \triangle AEC$.
$\therefore \frac{AD}{AE}=\frac{AB}{AC}$,即$\frac{AD}{AB}=\frac{AE}{AC}$.
在$\triangle DAE$和$\triangle BAC$中,
$\because \angle DAE=\angle BAC$,$\frac{AD}{AB}=\frac{AE}{AC}$,
$\therefore \triangle DAE \backsim \triangle BAC$.

8. 已知：如图,在$\triangle ABC$中,点$D$,$E分别在边BC$,$AC$上,点$F在DE$的延长线上,$AD= AF$,$AE\cdot CE= DE\cdot EF$。
(1)求证：$\triangle ADE\backsim\triangle ACD$；
(2)如果$AE\cdot BD= EF\cdot AF$,求证：$AB= AC$。

答案：
(1)证明:$\because AD=AF$,$\therefore \angle ADF=\angle F$.
$\because AE\cdot CE=DE\cdot EF$,$\therefore \frac{AE}{DE}=\frac{EF}{CE}$.
又$\because \angle AEF=\angle DEC$,$\therefore \triangle AEF \backsim \triangle DEC$.
$\therefore \angle F=\angle C$,$\therefore \angle ADF=\angle C$.
又$\because \angle DAE=\angle CAD$,$\therefore \triangle ADE \backsim \triangle ACD$.
(2)证明:$\because AE\cdot BD=EF\cdot AF$,
$\therefore \frac{AE}{AF}=\frac{EF}{BD}$.
$\because AD=AF$,$\therefore \frac{AE}{AD}=\frac{EF}{BD}$.
$\because \angle AEF=\angle EAD+\angle ADE$,$\angle ADB=$$\angle EAD+\angle C$,
$\therefore \angle AEF=\angle ADB$.$\therefore \triangle AEF \backsim \triangle ADB$.
$\therefore \angle F=\angle B$.$\therefore \angle C=\angle B$.$\therefore AB=AC$.

9. 已知：如图,$\triangle ADE的顶点E在\triangle ABC的边BC$上,$DE与AB相交于点F$,$AE^{2}= AF\cdot AB$,$\angle DAF= \angle EAC$。求证：
(1)$\triangle ADE\backsim\triangle ACB$；
(2)$\frac{DF}{DE}= \frac{CE}{CB}$。

答案：
(1)证明:$\because AE^{2}=AF\cdot AB$,$\therefore \frac{AE}{AB}=\frac{AF}{AE}$.
又$\because \angle EAF=\angle BAE$,$\therefore \triangle AEF \backsim \triangle ABE$.
$\therefore \angle AEF=\angle B$.
$\because \angle DAF=\angle EAC$,$\therefore \angle DAE=\angle BAC$.
$\therefore \triangle ADE \backsim \triangle ACB$.
(2)证明:$\because \triangle ADE \backsim \triangle ACB$,
$\therefore \frac{DE}{CB}=\frac{AD}{AC}$,$\angle D=\angle C$.
$\because \angle DAF=\angle EAC$,$\therefore \triangle ADF \backsim \triangle ACE$.
$\therefore \frac{AD}{AC}=\frac{DF}{CE}$.$\therefore \frac{DE}{BC}=\frac{DF}{CE}$.$\therefore \frac{DF}{DE}=\frac{CE}{CB}$.