《2025年暑期升级训练八年级数学浙教版浙江教育出版社》

1. 如图,若$AB= AC$,$BD= CD$,$∠A= 80^{\circ}$,$∠BDC= 120^{\circ}$,求$∠B$的度数.

答案：
如图, 连结AD并延长至点F. 在$\triangle ABD$和$\triangle ACD$中,$\because \left\{\begin{array}{l} AB=AC,\\ AD=AD,\\ BD=CD,\end{array}\right. $$\therefore \triangle ABD\cong \triangle ACD.$$\therefore ∠BAD=∠CAD,∠B=∠C.$$\because ∠BDF=∠B+∠BAD,∠CDF=∠C+∠CAD,$$\therefore ∠BDF+∠CDF=∠B+∠BAD+∠C+∠CAD.$$\therefore ∠BDC=∠B+∠C+∠BAC.$$\because ∠BAC=80^{\circ },∠BDC=120^{\circ },$$\therefore ∠B=20^{\circ }.$

2. 如图,$D为锐角∠ABC$内一点,点$M在边BA$上,点$N在边BC$上,$DM= DN$,$∠BMD+∠BND= 180^{\circ}$. 求证：$BD平分∠ABC$.

答案：
如图, 过点D分别作$DE⊥BA,DF⊥BC$, 垂足分别为E,F. $\because ∠BMD+∠BND=180^{\circ },∠BMD+∠EMD=180^{\circ },$$\therefore ∠EMD=∠BND$, 即$∠EMD=∠FND.$$\because DE⊥BA,DF⊥BC,$$\therefore ∠DEM=∠DFN=90^{\circ }$. 又$\because DM=DN,$$\therefore \triangle DEM\cong \triangle DFN.$$\therefore DE=DF$. 又$\because DE⊥BA,DF⊥BC,$$\therefore BD$平分$∠ABC$.

3. 如图,在$Rt△ABC$中,$∠ACB= 90^{\circ}$,$AC= CB$,$D为BC$的中点,$CE⊥AD于点E$,其延长线交$AB于点F$,连结$DF$. 求证：$∠ADC= ∠BDF$.

答案：
如图, 过点B作$BG⊥BC$, 交CF的延长线于点G. $\because ∠ACB=90^{\circ },$$\therefore ∠2+∠ACF=90^{\circ }$. $\because CE⊥AD,$$\therefore ∠AEC=90^{\circ }$. $\therefore ∠1+∠ACF=180^{\circ }-∠AEC=90^{\circ }$. $\therefore ∠1=∠2$. 在$\triangle ACD$和$\triangle CBG$中,$\because \left\{\begin{array}{l} ∠1=∠2,\\ AC=CB,\\ ∠ACD=∠CBG=90^{\circ },\end{array}\right. $$\therefore \triangle ACD\cong \triangle CBG.$$\therefore ∠ADC=∠G,CD=BG$. $\because D$为$BC$的中点,$\therefore CD=BD$. $\therefore BD=BG$. $\because ∠ACB=90^{\circ },AC=CB,$$\therefore ∠ABC=45^{\circ }$, 即$∠DBF=45^{\circ }$. 又$\because ∠DBG=90^{\circ },$$\therefore ∠GBF=∠DBG-∠DBF=90^{\circ }-45^{\circ }=45^{\circ }.$$\therefore ∠DBF=∠GBF$. 在$\triangle BDF$和$\triangle BGF$中,$\because \left\{\begin{array}{l} BD=BG,\\ ∠DBF=∠GBF,\\ BF=BF,\end{array}\right. $$\therefore \triangle BDF\cong \triangle BGF$. $\therefore ∠BDF=∠G$. $\therefore ∠ADC=∠BDF$.