《2025年学霸题中题八年级数学上册人教版》

6. 如图,AD是$\triangle ABC$的角平分线,$DF⊥AB$,垂足为F,$DE=DG,\triangle ADG$和$\triangle AED$的面积分别为50和39,则$\triangle EDF$的面积为____.

答案：
5.5 解析：如图，作$DM = DE$交$AC$于点$M$，过点$D$作$DN\perp AC$于点$N$。$\because DE = DG$，$\therefore DM = DG$。在$Rt\triangle DNM$和$Rt\triangle DNG$中，$DM = DG$，$DN = DN$，$\therefore Rt\triangle DNM\cong Rt\triangle DNG(HL)$。$\because AD$是$\triangle ABC$的角平分线，$DF\perp AB$，$\therefore$在$\triangle AFD$和$\triangle AND$中，$\begin{cases}\angle AFD = \angle AND\\\angle DAF = \angle DAN\\AD = AD\end{cases}$，$\therefore \triangle AFD\cong \triangle AND(AAS)$，$\therefore DF = DN$。
在$Rt\triangle DEF$和$Rt\triangle DMN$中，$DF = DN$，$DE = DM$，$\therefore Rt\triangle DEF\cong Rt\triangle DMN(HL)$，$\therefore S_{\triangle AMD}=S_{\triangle AED}$。$\because \triangle ADG$和$\triangle AED$的面积分别为 50 和 39，$\therefore S_{\triangle MDG}=S_{\triangle ADG}-S_{\triangle ADM}=50 - 39 = 11$，$\therefore S_{\triangle DNM}=S_{\triangle DNG}=S_{\triangle EDF}=\frac{1}{2}S_{\triangle MDG}=\frac{1}{2}\times11 = 5.5$。

归纳总结：角平分线模型
①如图①，过角平分线上一点作角两边的垂线段，构造 AAS 型全等；
②如图②，在角的两边上截长或补短，在角平分线两侧构造 SAS 型全等；
③如图③，当垂线段与角平分线垂直时，延长垂线段，构造 ASA 型全等(常与等腰三角形结合)；
④如图④，过角平分线上一点作角的一边的平行线，构造等腰三角形。

7. 已知$\triangle ABC$中,$∠A=60^{\circ },BD,CE$分别平分$∠ABC$和$∠ACB$,BD,CE交于点O,试判断BE,CD,BC的数量关系,并说明理由.

答案：
$BE + CD = BC$。理由如下：如图，在$BC$上截取$BF = BE$，连接$OF$，$AO$，$\because BD$平分$\angle ABC$，$\therefore \angle OBE = \angle OBF$。又$\because OB = OB$，$\therefore \triangle BEO\cong \triangle BFO(SAS)$，$\therefore \angle 1 = \angle 2$。$\because \angle BAC = 60^{\circ}$，$\therefore \angle BOC = 180^{\circ}-\frac{1}{2}\times(\angle ABC + \angle ACB)=180^{\circ}-\frac{1}{2}\times(180^{\circ}-60^{\circ}) = 120^{\circ}$，$\therefore \angle DOE = 120^{\circ}$，$\therefore \angle BAC + \angle DOE = 180^{\circ}$。$\because \angle AEO + \angle AOE + \angle EAO + \angle DAO + \angle DOA + \angle ADO = 180^{\circ}+180^{\circ}=360^{\circ}$，$\therefore \angle AEO + \angle ADO = 180^{\circ}$，$\therefore \angle 1 + \angle 4 = 180^{\circ}$。$\because \angle 2 + \angle 3 = 180^{\circ}$，且$\angle 1 = \angle 2$，$\therefore \angle 3 = \angle 4$。又$\because \angle DCO = \angle FCO$，$CO = CO$，$\therefore \triangle CDO\cong \triangle CFO(AAS)$，$\therefore CD = CF$，$\therefore BC = BF + CF = BE + CD$。

8. 已知点C是$∠MAN$平分线上一点,$∠BCD$的两边CB,CD分别与射线AM,AN相交于B,D两点,且$∠ABC+∠ADC=180^{\circ }$.过点C作$CE⊥AB$,垂足为E.
(1)如图①,当点E在线段AB上时,求证:$BC=DC$;
(2)如图②,当点E在线段AB的延长线上时,探究线段AB,AD与BE之间的数量关系;
(3)如图③,在(2)的条件下,若$∠MAN=60^{\circ }$,连接BD,作$∠ABD$的平分线BF交AD于点F,交AC于点O,连接DO并延长交AB于点G.若$BG=1,DF=2$,求线段DB的长.

答案：

(1)如图①，过点$C$作$CF\perp AD$，垂足为$F$。$\because AC$平分$\angle MAN$，$CE\perp AB$，$CF\perp AD$，$\therefore CE = CF$。$\because \angle CBE + \angle ADC = 180^{\circ}$，$\angle CDF + \angle ADC = 180^{\circ}$，$\therefore \angle CBE = \angle CDF$。在$\triangle BCE$和$\triangle DCF$中，$\begin{cases}\angle CBE = \angle CDF\\\angle CEB = \angle CFD = 90^{\circ}\\CE = CF\end{cases}$，$\therefore \triangle BCE\cong \triangle DCF(AAS)$，$\therefore BC = DC$。



(2)$AD - AB = 2BE$，理由如下：如图②，过点$C$作$CF\perp AD$，垂足为$F$。$\because AC$平分$\angle MAN$，$CE\perp AB$，$CF\perp AD$，$\therefore CE = CF$。又$\because AC = AC$，$\therefore Rt\triangle ACF\cong Rt\triangle ACE(HL)$，$\therefore AF = AE$。$\because \angle ABC + \angle ADC = 180^{\circ}$，$\angle ABC + \angle CBE = 180^{\circ}$，$\therefore \angle CDF = \angle CBE$。在$\triangle BCE$和$\triangle DCF$中，$\begin{cases}\angle CBE = \angle CDF\\\angle CEB = \angle CFD = 90^{\circ}\\CE = CF\end{cases}$，$\therefore \triangle BCE\cong \triangle DCF(AAS)$，$\therefore DF = BE$，$\therefore AD = AF + DF = AE + DF = AB + BE + DF = AB + 2BE$，$\therefore AD - AB = 2BE$。
(3)如图③，在$BD$上截取$BH = BG$，连接$OH$。$\because BH = BG$，$\angle OBH = \angle OBG$，$OB = OB$，$\therefore \triangle OBH\cong \triangle OBG(SAS)$，$\therefore \angle OHB = \angle OGB$。$\because AO$是$\angle MAN$的平分线，$BO$是$\angle ABD$的平分线，$\therefore$点$O$到$AD$，$AB$，$BD$的距离相等，$\therefore \angle ODH = \angle ODF$。$\because \angle OHB = \angle ODH + \angle DOH$，$\angle OGB = \angle ODF + \angle DAB$，$\therefore \angle DOH = \angle DAB = 60^{\circ}$，$\therefore \angle GOH = 120^{\circ}$，$\therefore \angle BOG = \angle BOH = 60^{\circ}$，$\therefore \angle DOF = \angle BOG = 60^{\circ}$，$\therefore \angle DOH = \angle DOF$。在$\triangle ODH$和$\triangle ODF$中，$\begin{cases}\angle DOH = \angle DOF\\OD = OD\\\angle ODH = \angle ODF\end{cases}$，$\therefore \triangle ODH\cong \triangle ODF(ASA)$，$\therefore DH = DF$，$\therefore DB = DH + BH = DF + BG = 2 + 1 = 3$。