《2025年启东中学作业本八年级数学上册苏科版连淮专版》

6. 如图,在$\triangle ABC$中,$∠ABC= 2∠C$,$BQ和AP分别为∠ABC和∠BAC$的平分线,若$\triangle ABQ$的周长为18,$BP= 4$,求$AB$的长.

答案：
解:$\because BQ$平分$∠ABC,\therefore ∠CBQ = \frac{1}{2}∠ABC.$
$\because ∠ABC = 2∠C,\therefore ∠CBQ = ∠C,\therefore BQ = CQ,$
$\therefore BQ + AQ = CQ + AQ = AC$①.
过点P作$PD// BQ$交CQ于点D,如答图,

则$∠CPD = ∠CBQ = ∠C,∠ADP = ∠AQB,$
$\therefore PD = CD.$
$\because ∠AQB = ∠C + ∠CBQ = 2∠C,\therefore ∠ABC = ∠ADP.$
$\because AP$平分$∠BAC,\therefore ∠BAP = ∠CAP.$
在$\triangle ABP$和$\triangle ADP$中,$\left\{\begin{array}{l} ∠ABP=∠ADP,\\ ∠BAP=∠DAP,\\ AP=AP,\end{array}\right. $
$\therefore \triangle ABP\cong \triangle ADP(AAS),\therefore AB = AD,BP = PD,$
$\therefore AB + BP = AD + PD = AD + CD = AC$②,
由①②得$BQ + AQ = AB + BP.$
$\because \triangle ABQ$的周长为18,$BP = 4,\therefore AB + BQ + AQ = AB + BP + AB = 2AB + 4 = 18,\therefore AB = 7.$

7. 如图,在$\triangle ABC$中,$∠B= 60^{\circ}$,$D$,$E分别为AB$,$BC$上的点,且$AE$,$CD交于点F$. 若$AE$,$CD分别为\triangle ABC$的角平分线.
(1)求证：$∠AFC= 120^{\circ}$；
(2)若$AD= 6$,$CE= 4$,求$AC$的长.

答案：

(1)证明:$\because AE,CD$分别为$\triangle ABC$的角平分线,
$\therefore ∠FAC = \frac{1}{2}∠BAC,∠FCA = \frac{1}{2}∠BCA.$
$\because ∠B = 60^{\circ },\therefore ∠BAC + ∠BCA = 120^{\circ },$
$\therefore ∠AFC = 180^{\circ } - ∠FAC - ∠FCA = 180^{\circ } - \frac{1}{2}(∠BAC + ∠BCA) = 180^{\circ } - \frac{1}{2}×120^{\circ } = 120^{\circ }.$
(2)解:如答图，在AC上截取$AG = AD$，连接FG.
$\because AE,CD$分别为$\triangle ABC$的角平分线,
$\therefore ∠FAG = ∠FAD,∠FCG = ∠FCE.$
$\because ∠AFC = 120^{\circ },\therefore ∠AFD = ∠CFE = 60^{\circ }.$
在$\triangle ADF$和$\triangle AGF$中,$\left\{\begin{array}{l} AD=AG,\\ ∠FAD=∠FAG,\\ AF=AF,\end{array}\right. $
$\therefore \triangle ADF\cong \triangle AGF(SAS),\therefore ∠AFD = ∠AFG = 60^{\circ },$
$\therefore ∠GFC = ∠AFC - ∠AFG = 60^{\circ },$
$\therefore ∠GFC = ∠CFE.$
在$\triangle CGF$和$\triangle CEF$中,$\left\{\begin{array}{l} ∠GFC=∠EFC,\\ CF=CF,\\ ∠FCG=∠FCE,\end{array}\right. $
$\therefore \triangle CGF\cong \triangle CEF(ASA),\therefore CG = CE,$
$\therefore AC = AG + CG = AD + CE = 10.$

8. 如图①,$AD平分∠BAC$,$∠B+∠C= 180^{\circ}$,$∠B= 90^{\circ}$,易知$DB= DC$.
(1)如图②,$AD平分∠BAC$,$∠ABD+∠ACD= 180^{\circ}$,$∠ABD＜90^{\circ}$. 求证：$DB= DC$；
(2)如图③,在四边形$ABDC$中,$∠B= 45^{\circ}$,$∠C= 135^{\circ}$,$DB= DC$,$DE⊥AB$. 求证：$AB-AC= 2BE$.

答案：
证明:
(1)如答图①,过点D作$DE⊥AB$于点E,$DF⊥AC$，交AC的延长线于点F.
$\because AD$平分$∠BAC,\therefore ∠DAC = ∠DAB.$
在$\triangle DAF$和$\triangle DAE$中,$\left\{\begin{array}{l} ∠DAF=∠DAE,\\ ∠DFA=∠DEA,\\ AD=AD,\end{array}\right. $
$\therefore \triangle DAF\cong \triangle DAE(AAS),\therefore DF = DE.$
$\because ∠ABD + ∠ACD = 180^{\circ },∠ACD + ∠FCD = 180^{\circ },$
$\therefore ∠ABD = ∠FCD.$
在$\triangle DFC$和$\triangle DEB$中,$\left\{\begin{array}{l} ∠DFC=∠DEB,\\ ∠FCD=∠EBD,\\ DF=DE,\end{array}\right. $
$\therefore \triangle DFC\cong \triangle DEB(AAS),\therefore DB = DC.$
(2)如答图②,连接AD,过点D作$DF⊥AC$，交AC的延长线于点F.
$\because ∠ACD = 135^{\circ },\therefore ∠FCD = 180^{\circ } - ∠ACD = 45^{\circ }.$
$\because ∠B = 45^{\circ },\therefore ∠FCD = ∠B.$
在$\triangle DFC$和$\triangle DEB$中,$\left\{\begin{array}{l} ∠DFC=∠DEB=90^{\circ },\\ ∠FCD=∠B,\\ DC=DB,\end{array}\right. $
$\therefore \triangle DFC\cong \triangle DEB(AAS),\therefore DF = DE,CF = BE.$
在$Rt\triangle ADF$和$Rt\triangle ADE$中,$\left\{\begin{array}{l} AD=AD,\\ DF=DE,\end{array}\right. $
$\therefore Rt\triangle ADF\cong Rt\triangle ADE(HL),\therefore AF = AE,$
$\therefore AB = AE + BE = AC + CF + BE = AC + 2BE,$
$\therefore AB - AC = 2BE.$