《2025年暑期升级训练八年级数学浙教版浙江教育出版社》

8. 如图,$AD是△ABC$的外角平分线. 下列结论中,一定正确的是 (D)

A.$AD+BC= AB+CD$
B.$AB+AC= DB+DC$
C.$AD+BC＜AB+CD$
D.$DB+DC＞AB+AC$

答案： D 解析: 在AP上截取$AF=AC$, 连结DF. $\because AD$是$\triangle ABC$的外角平分线, $\therefore ∠DAF=∠DAC$. 又$\because AF=AC,AD=AD,$$\therefore \triangle DAF\cong \triangle DAC.$$\therefore DF=DC.$$\because DB+DF＞BF,BF=AB+AF,$$\therefore DB+DC＞AB+AC$.

9. 如图,$AB// CD$,$BE平分∠ABC$,$CE平分∠BCD$,点$E在AD$上. 求证：$BC= AB+CD$.

答案：
如图, 在BC上截取$FB=AB$, 连结EF. $\because BE$平分$∠ABC,CE$平分$∠BCD,$$\therefore ∠ABE=∠FBE,∠FCE=∠DCE$. 在$\triangle ABE$和$\triangle FBE$中,$\because \left\{\begin{array}{l} AB=FB,\\ ∠ABE=∠FBE,\\ BE=BE,\end{array}\right. $$\therefore \triangle ABE\cong \triangle FBE.$$\therefore ∠A=∠BFE.$$\because AB// CD,$$\therefore ∠A+∠D=180^{\circ }.$$\therefore ∠BFE+∠D=180^{\circ }.$$\because ∠BFE+∠CFE=180^{\circ },$$\therefore ∠CFE=∠D$. 在$\triangle FCE$和$\triangle DCE$中,$\because \left\{\begin{array}{l} ∠CFE=∠D,\\ ∠FCE=∠DCE,\\ CE=CE,\end{array}\right. $$\therefore \triangle FCE\cong \triangle DCE.$$\therefore CF=CD.$$\therefore BC=FB+CF=AB+CD$.

10. 如图,在$△ABC$中,$AB＞AC$,$∠1= ∠2$,$P为AD$上任意一点. 求证：$AB-AC＞PB-PC$.

答案：
如图, 延长AC至点M, 使$AM=AB$, 连结PM. 在$\triangle ABP$和$\triangle AMP$中,$\because AB=AM,∠1=∠2,AP=AP,$$\therefore \triangle ABP\cong \triangle AMP.$$\therefore PB=PM$. 在$\triangle PCM$中, 根据三角形的三边关系, 得$CM＞PM-PC$,$\therefore AM-AC＞PB-PC.$$\therefore AB-AC＞PB-PC$.

11. 如图,在$Rt△ABC$中,$∠BAC= 90^{\circ}$,$∠B= 60^{\circ}$,$AD$,$CE分别平分∠BAC$,$∠ACB$,且相交于点$O$.
(1)求$∠AOE$的度数；
(2)求证：$AC= AE+CD$.

答案：

 
(1) $\because ∠BAC=90^{\circ },∠B=60^{\circ },$$\therefore ∠ACB=180^{\circ }-90^{\circ }-60^{\circ }=30^{\circ }$. $\because AD,CE$分别平分$∠BAC,∠ACB,$$\therefore ∠CAO=\frac {1}{2}∠BAC=45^{\circ },∠ACO=\frac {1}{2}∠ACB=15^{\circ }.$$\therefore ∠AOE=∠CAO+∠ACO=45^{\circ }+15^{\circ }=60^{\circ }$.

(2) 如图, 在AC上截取$AF=AE$, 连结OF. $\because AD$平分$∠BAC,$$\therefore ∠EAO=∠FAO$. 在$\triangle AOE$和$\triangle AOF$中,$\because \left\{\begin{array}{l} AE=AF,\\ ∠EAO=∠FAO,\\ AO=AO,\end{array}\right. $$\therefore \triangle AOE\cong \triangle AOF.$$\therefore ∠AOE=∠AOF$. 由
(1), 知$∠AOE=60^{\circ },$$\therefore ∠AOF=60^{\circ },∠COD=60^{\circ }.$$\therefore ∠COF=180^{\circ }-∠AOF-∠COD=60^{\circ }.$$\therefore ∠COF=∠COD$. $\because CE$平分$∠ACB,$$\therefore ∠FCO=∠DCO$. 在$\triangle COF$和$\triangle COD$中,$\because \left\{\begin{array}{l} ∠COF=∠COD,\\ CO=CO,\\ ∠FCO=∠DCO,\end{array}\right. $$\therefore \triangle COF\cong \triangle COD.$$\therefore CF=CD.$$\therefore AC=AF+CF=AE+CD$.

 
(1) $\because ∠BAC=90^{\circ },∠B=60^{\circ },$$\therefore ∠ACB=180^{\circ }-90^{\circ }-60^{\circ }=30^{\circ }$. $\because AD,CE$分别平分$∠BAC,∠ACB,$$\therefore ∠CAO=\frac {1}{2}∠BAC=45^{\circ },∠ACO=\frac {1}{2}∠ACB=15^{\circ }.$$\therefore ∠AOE=∠CAO+∠ACO=45^{\circ }+15^{\circ }=60^{\circ }$.

(2) 如图, 在AC上截取$AF=AE$, 连结OF. $\because AD$平分$∠BAC,$$\therefore ∠EAO=∠FAO$. 在$\triangle AOE$和$\triangle AOF$中,$\because \left\{\begin{array}{l} AE=AF,\\ ∠EAO=∠FAO,\\ AO=AO,\end{array}\right. $$\therefore \triangle AOE\cong \triangle AOF.$$\therefore ∠AOE=∠AOF$. 由
(1), 知$∠AOE=60^{\circ },$$\therefore ∠AOF=60^{\circ },∠COD=60^{\circ }.$$\therefore ∠COF=180^{\circ }-∠AOF-∠COD=60^{\circ }.$$\therefore ∠COF=∠COD$. $\because CE$平分$∠ACB,$$\therefore ∠FCO=∠DCO$. 在$\triangle COF$和$\triangle COD$中,$\because \left\{\begin{array}{l} ∠COF=∠COD,\\ CO=CO,\\ ∠FCO=∠DCO,\end{array}\right. $$\therefore \triangle COF\cong \triangle COD.$$\therefore CF=CD.$$\therefore AC=AF+CF=AE+CD$.