《2025年华东师大版一课一练九年级数学全一册沪教版54制》

13 在$\triangle ABC$中,$\angle ABC= \angle ACB$,$B为AE$的中点,$AD= BD$。
求证：(1)$\triangle ACE\backsim\triangle ADC$；
(2)$CE= 2DC$。

答案：
(1) $\because\angle ABC=\angle ACB$, B为AE的中点,$\therefore AB=AC$, $AB=BE$。$\because AD=BD$,$\therefore\frac{AD}{AB}=\frac{AD}{AC}=\frac{1}{2}$。$\because AB=BE=AC$,$\therefore\frac{AC}{AE}=\frac{1}{2}$。$\therefore\frac{AD}{AC}=\frac{AC}{AE}$。$\because\angle A=\angle A$,$\therefore\triangle ACE\backsim\triangle ADC$。
(2) 由
(1)得$\triangle ACE\backsim\triangle ADC$,$\therefore\frac{DC}{CE}=\frac{AC}{AE}=\frac{1}{2}$,即$CE=2DC$。

14 已知点$D在\triangle ABC$的外部,$AD// BC$,点$E在边AB$上,$AB\cdot AD= BC\cdot AE$。
(1)求证：$\angle BAC= \angle AED$；
(2)在边$AC上取一点F$,如果$\angle AFE= \angle D$,求证：$\frac{AD}{BC}= \frac{AF}{AC}$。

答案：
(1) $\because AD// BC$,$\therefore\angle B=\angle DAE$。$\because AB\cdot AD=BC\cdot AE$,$\therefore\frac{AB}{BC}=\frac{AE}{AD}$,$\therefore\triangle CBA\backsim\triangle DAE$,$\therefore\angle BAC=\angle AED$。
(2) 由
(1)得$\triangle DAE\backsim\triangle CBA$,$\therefore\angle D=\angle C$,$\frac{AD}{BC}=\frac{DE}{AC}$。$\because\angle AFE=\angle D$,$\therefore\angle AFE=\angle C$。$\therefore EF// BC$,$\because AD// BC$,$\therefore EF// AD$。$\because\angle BAC=\angle AED$,$\therefore DE// AC$。$\therefore$四边形ADEF是平行四边形。$\therefore DE=AF$。$\therefore\frac{AD}{BC}=\frac{AF}{AC}$。

15 已知：$\triangle ADE的顶点E在\triangle ABC的边BC$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) $\because AE^{2}=AF\cdot AB$,$\therefore\frac{AF}{AE}=\frac{AE}{AB}$。又$\because\angle FAE=\angle EAB$,$\therefore\triangle AFE\backsim\triangle AEB$。$\therefore\angle AEF=\angle B$。$\because\angle DAF=\angle EAC$,$\therefore\angle DAE=\angle CAB$。$\therefore\triangle ADE\backsim\triangle ACB$。
(2) $\because\triangle ADE\backsim\triangle ACB$,$\therefore\angle D=\angle C$,$\frac{AD}{AC}=\frac{DE}{CB}$。$\because\angle DAF=\angle EAC$,$\therefore\triangle ADF\backsim\triangle ACE$。$\therefore\frac{AD}{AC}=\frac{DF}{CE}$。$\therefore\frac{DF}{CE}=\frac{DE}{CB}$。$\therefore\frac{DF}{DE}=\frac{CE}{CB}$。