《2025年假期新思维七年级数学北师大版》

21.(9分)如图所示,在$\triangle ABC$中,$\angle C= 90^{\circ}$,$AD是\angle BAC$的平分线,$DE\perp AB交AB于E$,$F在AC$上,$BD= DF$。
求证：(1)$CF= EB$；(2)$AB= AF+2EB$。

答案： 证明:
(1) $\because AD$是$\angle BAC$的平分线, $DE\perp AB,\angle C = 90^{\circ}$,
$\therefore DE = DC$, 又$\because BD = DF,\therefore Rt\triangle CDF\cong Rt\triangle EDB(HL),\therefore CF = EB$.
(2) $\because AD$是$\angle BAC$的平分线, $DE\perp AB,\angle C = 90^{\circ},\therefore \triangle ADC\cong\triangle ADE,\therefore AC = AE,\therefore AB$
$= AE + BE = AC + EB = AF + CF + EB = AF + 2EB$.

22.(9分)(阅读理解题)如图所示,$CE\perp AB于点E$,$BD\perp AC于点D$,$BD$,$CE交于点O$,且$AO平分\angle BAC$。
(1)图中有多少对全等三角形？请一一列举出来(不必说明理由)；
(2)小明说：欲证$BE= CD$,可先证明$\triangle AOE\cong\triangle AOD得到AE= AD$,再证明$\triangle ADB\cong\triangle AEC得到AB= AC$,然后利用等式的性质得到$BE= CD$,请问他的说法正确吗？如果正确,请按照他的说法写出推导过程,如果不正确,请说明理由；
(3)要得到$BE= CD$,你还有其他思路吗？

答案： 解:
(1) 有$4$对, 分别是$\triangle AOE\cong\triangle AOD,\triangle BOE\cong\triangle COD,\triangle AOB\cong\triangle AOC,\triangle ABD$
$\cong\triangle ACE$.
(2) 小明的说法正确.
$\because CE\perp AB$于点$E,BD\perp AC$于点$D,\therefore \angle AEO=\angle ADO = 90^{\circ},\because AO$平分$\angle BAC,\therefore$
$\angle OAE=\angle OAD$, 在$\triangle AOE$和$\triangle AOD$中, $\because \begin{cases}\angle AEO=\angle ADO\\\angle OAE=\angle OAD\\AO = AO\end{cases},\therefore \triangle AOE\cong\triangle AOD$
$(AAS).\therefore AE = AD$. 在$\triangle ADB$和$\triangle AEC$中, $\because \begin{cases}\angle ADB=\angle AEC\\AD = AE\\\angle BAD=\angle CAE\end{cases},\therefore \triangle ADB\cong\triangle AEC$
$(ASA).\therefore AB = AC,\therefore AB - AE = AC - AD$, 即$BE = CD$.
(3) 可先证$\triangle AOE\cong\triangle AOD$得到$OE = OD$, 再证
$\triangle BOE\cong\triangle COD$得到$BE = CD$.

23.(12分)如图1,在$\triangle ABC$中,$\angle ABC= 45^{\circ}$,高线$AD$、$DE相交于点F$。
(1)判断$BF与AC$的数量关系并说明理由；
(2)如图2,将$\triangle ACD沿线段AD$对折,点$C落在BD上的点M$,$AM与BE相交于点N$,当$DE// AM$时,判断$NE与AC$的数量关系并说明理由。

答案： 解:
(1) $BF = AC$, 理由是:
如图1, $\because AD\perp BC,BE\perp AC,\therefore \angle ADB=\angle AEF = 90^{\circ},\because \angle ABC = 45^{\circ},\therefore \triangle ABD$是等腰
直角三角形, $\therefore AD = BD,\because \angle AFE=\angle BFD,\therefore \angle DAC=\angle EBC$, 在$\triangle ADC$和$\triangle BDF$中,
$\because \begin{cases}\angle DAC=\angle DBF\\\angle ADC=\angle BDF = 90^{\circ}\\AD = BD\end{cases},\therefore \triangle ADC\cong\triangle BDF(AAS),\therefore BF = AC$;
(2) $NE=\frac{1}{2}AC$, 理由是:
如图2, 由折叠得: $MD = DC,\because DE// AM,\therefore AE = EC,\because BE\perp AC,\therefore AB = BC,\therefore \angle ABE$
$\angle CBE$, 由
(1)得: $\triangle ADC\cong\triangle BDF,\because \triangle ADC\cong\triangle ADM,\therefore \triangle BDF\cong\triangle ADM,\therefore \angle DBF$
$\angle MAD,\because \angle DBA=\angle BAD = 45^{\circ},\therefore \angle DBA-\angle DBF=\angle BAD-\angle MAD$, 即$\angle ABE$
$\angle BAN,\because \angle ANE=\angle ABE+\angle BAN = 2\angle ABE,\angle NAE = 2\angle NAD = 2\angle CBE,\therefore \angle ANE$
$\angle NAE = 45^{\circ},\therefore AE = EN,\therefore EN=\frac{1}{2}AC$.