《2025年零障碍导教导学案九年级数学全一册人教版》

5. 如图,在等边三角形$ABC$中,$D为BC$上一点,$∠EDF= 60^{\circ }$,证明图中一对相似三角形.

答案： 解：$\triangle EBD\backsim \triangle DCF$. 证明如下：
$\because \triangle ABC$是等边三角形，
$\therefore \angle B=\angle C.$
$\because \angle 2=60^{\circ }+\angle 1-\angle B=\angle 1,$
$\therefore \triangle EBD\backsim \triangle DCF.$

6. 如图,在矩形$ABCD$中,$E为BC$上的一点,作$∠AEF= 90^{\circ },EF交CD于点F$,指出图中一对相似三角形并证明.
$\triangle ABE\backsim \triangle ECF$
证明如下：
$\because \angle B=\angle AEF=90^{\circ },$
$\therefore \angle BAE+\angle AEB=\angle CEF+\angle AEB$
$=90^{\circ }.$
$\therefore \angle BAE=\angle CEF.$
又$\because \angle B=\angle C=90^{\circ },$
$\therefore \triangle ABE\backsim \triangle ECF.$

答案： 解：$\triangle ABE\backsim \triangle ECF$. 证明如下：
$\because \angle B=\angle AEF=90^{\circ },$
$\therefore \angle BAE+\angle AEB=\angle CEF+\angle AEB$
$=90^{\circ }.$
$\therefore \angle BAE=\angle CEF.$
又$\because \angle B=\angle C=90^{\circ },$
$\therefore \triangle ABE\backsim \triangle ECF.$

7. (2024·斗门区模拟)如图,已知$∠B= ∠ACD$.
(1)求证:$△ABC\backsim △ACD$;
证明：$\because \angle B=\angle ACD,\angle A=\angle A,$
$\therefore \triangle ABC\backsim \triangle ACD.$
(2)当$AD= 2,AB= 3$时,求$AC$的长.
解：$\because \triangle ABC\backsim \triangle ACD,$
$\therefore \frac {AC}{AD}=\frac {AB}{AC}$，即$\frac {AC}{2}=\frac {3}{AC}$
$\therefore AC=$$\sqrt{6}$.

答案： 证明：$\because \angle B=\angle ACD,\angle A=\angle A,$
$\therefore \triangle ABC\backsim \triangle ACD.$
(2) 解：$\because \triangle ABC\backsim \triangle ACD,$
$\therefore \frac {AC}{AD}=\frac {AB}{AC}$，即$\frac {AC}{2}=\frac {3}{AC}$
$\therefore AC=\sqrt {6}.$

8. (2024·东莞校级一模)如图,在平行四边形$ABCD$中,$E为边DC$上一点,$∠EAB= ∠CBE$.
(1)求证:$△ABE\backsim △BEC$;
证明：$\because$ 四边形$ABCD$是平行四边形，
$\therefore AB// CD.$
$\therefore \angle EBA=\angle CEB.$
又$\because \angle EAB=\angle CBE.$
$\therefore \triangle ABE\backsim \triangle BEC.$
(2)若$AB= 4,DE= 3$,求$BE$的长.
解：$\because$ 四边形$ABCD$是平行四边形，
$\therefore AB=DC=4.$
$\because DE=3,$
$\therefore CE=1.$
$\because \triangle ABE\backsim \triangle BEC,$
$\therefore \frac {AB}{BE}=\frac {BE}{EC}.$
$\therefore AB\cdot CE=BE^{2}=4×1=4.$
$\therefore BE=$2.

答案：
(1) 证明：$\because$ 四边形$ABCD$是平行四边形，
$\therefore AB// CD.$
$\therefore \angle EBA=\angle CEB.$
又$\because \angle EAB=\angle CBE.$
$\therefore \triangle ABE\backsim \triangle BEC.$
(2) 解：$\because$ 四边形$ABCD$是平行四边形，
$\therefore AB=DC=4.$
$\because DE=3,$
$\therefore CE=1.$
$\because \triangle ABE\backsim \triangle BEC,$
$\therefore \frac {AB}{BE}=\frac {BE}{EC}.$
$\therefore AB\cdot CE=BE^{2}=4×1=4.$
$\therefore BE=2.$

9. 如图,$△ABC内接于\odot O$,弦$CD= BC$,弦$AC与BD相交于点P$.
(1)求证:$△CAB\backsim △CBP$;
(2)若$BC= 4\sqrt {3},CP= 6$,求$AP$的长.
(1) 证明：$\because \angle BAC$与$\angle BDC$为同弦$BC$所对的圆周角，
$\therefore \angle BAC=\angle BDC.$
$\because CD=BC,$
$\therefore \angle BDC=\angle DBC,$
即$\angle BAC=\angle DBC.$
又$\because \angle BCA=\angle PCB,$
$\therefore \triangle CAB\backsim \triangle CBP.$
(2) 解：由(1)得$\triangle CAB\backsim \triangle CBP,$
$\therefore \frac {AC}{BC}=\frac {BC}{PC}$，即$\frac {AC}{4\sqrt {3}}=\frac {4\sqrt {3}}{6},$
解得$AC=8.$
$\therefore AP=AC-CP=8-6=2.$

答案：
(1) 证明：$\because \angle BAC$与$\angle BDC$为同弦$BC$所对的圆周角，
$\therefore \angle BAC=\angle BDC.$
$\because CD=BC,$
$\therefore \angle BDC=\angle DBC,$
即$\angle BAC=\angle DBC.$
又$\because \angle BCA=\angle PCB,$
$\therefore \triangle CAB\backsim \triangle CBP.$
(2) 解：由
(1)得$\triangle CAB\backsim \triangle CBP,$
$\therefore \frac {AC}{BC}=\frac {BC}{PC}$，即$\frac {AC}{4\sqrt {3}}=\frac {4\sqrt {3}}{6},$
解得$AC=8.$
$\therefore AP=AC-CP=8-6=2.$

10. 如图,以$□ ABCD的边CD为直径作\odot O$,$\odot O与边BC相交于点F$,$\odot O的切线DE与边AB相交于点E$.
(1)求证:$△ADE\backsim △CDF$;
(2)若$\frac {CF}{AE}= \frac {2}{3},BC= 6$,求$\odot O$的面积为4π.

答案：
(1) 证明：$\because$ 四边形$ABCD$是平行四边形，
$\therefore \angle A=\angle DCF,AB// CD.$
$\because DC$是直径，
$\therefore \angle CFD=90^{\circ }.$
$\because DE$是$\odot O$的切线，
$\therefore \angle EDC=90^{\circ }.$
又$\because AB// CD,$
$\therefore \angle EDC=\angle AED=90^{\circ }.$
$\therefore \angle AED=\angle CFD.$
$\therefore \triangle ADE\backsim \triangle CDF.$
(2) 解：由
(1)得$\triangle ADE\backsim \triangle CDF,$
$\therefore \frac {CF}{AE}=\frac {DC}{AD}$，即$\frac {2}{3}=\frac {DC}{6},$
解得$DC=4.$
$\therefore DO=OC=2.$
$\therefore \odot O$的面积为$π×2^{2}=4π.$