《2025年名校题库八年级数学上册北师大版》

7. (石室联中)如图,在$\triangle ABC$中,$∠A = 45^{\circ}$,$∠ABC = 60^{\circ}$,$AB = 14$,$BD⊥AC$,$E$,$F$,$G分别是AD$,$BD$,$BC$上的动点,且$BF = DE$,则$\frac{\sqrt{2}}{2}EF + FG$的最小值为______。

答案：
$\frac{7\sqrt{3}}{2}$ [解析]如图，过点$D$作$DH \perp AB$于点$H$，连接$HE$，$HF$，$HG$。$\because BD \perp AC$，$\therefore \angle ADB = 90^{\circ}$。$\because \angle A = 45^{\circ}$，$\therefore \angle ABD = 45^{\circ}$，$\therefore \triangle ADB$为等腰直角三角形。$\because DH \perp AB$，$\therefore DH = AH = BH$，$\angle ADH = 45^{\circ}$。在$\triangle DEH$和$\triangle BFH$中，$\begin{cases}DE = BF\\\angle EDH = \angle FBH\\DH = BH\end{cases}$，$\therefore \triangle DEH \cong \triangle BFH(SAS)$，$\therefore HE = HF$，$\angle DHE = \angle BHF$，$\therefore \angle EHF = \angle DHE + \angle DHF = \angle BHF + \angle DHF = \angle DHB = 90^{\circ}$，$\therefore \triangle HEF$为等腰直角三角形，$\therefore HF = \frac{\sqrt{2}}{2}EF$，$\therefore \frac{\sqrt{2}}{2}EF + FG = HF + FG$。而$HF + FG \geq HG$（当且仅当$H$，$F$，$G$三点共线时取等号），$\therefore HF + FG$的最小值为$HG$的长。过点$H$作$HG' \perp BC$于点$G'$。当点$G$与点$G'$重合时，$HG$的长有最小值。在$Rt\triangle BHG'$中，$\angle HBG' = 60^{\circ}$，$BH = \frac{1}{2}AB = 7$，$\therefore BG' = \frac{1}{2}BH = \frac{7}{2}$，$\therefore HG' = \sqrt{3}BG = \frac{7\sqrt{3}}{2}$，$\therefore HF + FG$的最小值为$\frac{7\sqrt{3}}{2}$，即$\frac{\sqrt{2}}{2}EF + FG$的最小值为$\frac{7\sqrt{3}}{2}$。

8. (武侯区期末)如图,在$\triangle ABC$中,$AB = 4\sqrt{3}$,$∠ABC = 60^{\circ}$,$BD平分∠ABC交边AC于点D$,$AD = \frac{2}{3}CD$。在$BC边上取一点E$,连接$DE$,将线段$DE平移后得到线段BF$,连接$AF$,则线段$AF$的长度的最小值是______。

答案：
$\frac{48}{5}$ [解析]如图，过点$D$作$DM \perp BC$于点$M$，$DN \perp AB$于点$N$，过点$A$作$AG \perp BC$于点$G$，过点$F$作$FT \perp BC$于点$T$，连接$FG$，$EF$。$\because BD$平分$\angle ABC$，$DM \perp BC$，$DN \perp AB$，$\therefore DM = DN$，$\therefore \frac{S_{\triangle ABD}}{S_{\triangle BCD}} = \frac{AD}{CD} = \frac{\frac{1}{2}AB \cdot DN}{\frac{1}{2}BC \cdot DM} = \frac{AB}{BC}$。又$\because AD = \frac{2}{3}CD$，$\therefore \frac{AB}{BC} = \frac{2}{3}$。$\because AB = 4\sqrt{3}$，$\therefore BC = 6\sqrt{3}$。$\because AG \perp CB$，$\angle ABG = 60^{\circ}$，$\therefore \angle BAG = 30^{\circ}$，$\therefore BG = \frac{1}{2}AB = 2\sqrt{3}$，$\therefore AG = \sqrt{(4\sqrt{3})^{2}-(2\sqrt{3})^{2}} = 6$。$\because S_{\triangle ABC} = \frac{1}{2}BC \cdot AG = \frac{1}{2}AB \cdot DN + \frac{1}{2}BC \cdot DM$，$\therefore DM = DN = \frac{6\sqrt{3}×6}{4\sqrt{3} + 6\sqrt{3}} = \frac{18}{5}$。$\because DE // BF$，$\therefore \angle DEB = \angle EBF$。又$\because DE = BF$，$BE = EB$，$\therefore \triangle BED \cong \triangle EBF(SAS)$。$\because DM \perp BE$，$FT \perp BE$，$\therefore FT = DM = \frac{18}{5}$，$\therefore AF \geq AG + FT = 6 + \frac{18}{5} = \frac{48}{5}$，$\therefore$线段$AF$的长度的最小值为$\frac{48}{5}$。

9. (武侯区期末)如图,在$Rt\triangle ABC$中,$∠ACB = 90^{\circ}$,$∠BAC = 30^{\circ}$,延长$BC至点D$,使$CD = BC$,连接$AD$。
(1)求证：$\triangle ABD$是等边三角形。
(2)若$E为线段CD$的中点,且$AD = 4$,$P为线段AC$上一动点,连接$EP$,$BP$。
①求$EP + \frac{1}{2}AP$的最小值；
②求$2BP + AP$的最小值。

答案：

(1)证明：由题意易证$\triangle ABC \cong \triangle ADC$，$\therefore AB = AD$，$\angle DAC = \angle BAC = 30^{\circ}$，$\therefore \angle BAD = 60^{\circ}$，$\therefore \triangle ABD$为等边三角形。
(2)解：①如图1，过点$P$作$PH \perp AB$于点$H$。$\because \angle PAH = 30^{\circ}$，$\therefore PH = \frac{1}{2}AP$，$\therefore EP + \frac{1}{2}AP = EP + PH$。过点$E$作$EH' \perp AB$于点$H'$，$EH'$交$AC$于点$P'$，当点$P$，$H$分别与点$P'$，$H'$重合时，$EP + PH$最小，且最小值为$EH'$的长。又$\because AD = 4$，$CD = CB$，$EC = ED$，$\angle ABC = 60^{\circ}$，$\therefore BE = 3$，$\angle BEH' = 30^{\circ}$，$\therefore BH' = \frac{3}{2}$，$\therefore EH' = \frac{3\sqrt{3}}{2}$，$\therefore EP + \frac{1}{2}AP$的最小值为$\frac{3\sqrt{3}}{2}$。

②如图2，过点$B$作$BM \perp AD$于点$M$，交$AC$于点$P$，此时$BP + \frac{1}{2}AP$最小，$MP = \frac{1}{2}AP$，$\therefore BP + \frac{1}{2}AP = BP + MP = BM = 2\sqrt{3}$，$\therefore$此时$2BP + AP = 2(BP + \frac{1}{2}AP) = 2×2\sqrt{3} = 4\sqrt{3}$。即$2BP + AP$的最小值为$4\sqrt{3}$。