《2025年名校题库八年级数学上册北师大版》

12. (石室天府)如图,在$\triangle ABC$中,$∠ABC = ∠ACB = 30^{\circ}$,$AB = 2$,$D是BC$边上一动点,以$AD为边在AD的右侧作等边\triangle ADE$。
(1)当$AD与BC所夹的锐角为45^{\circ}$时,求$CD$的长。
(2)当点$D从点C向点B$移动过程中,
①求点$E$的移动路径的长度；
②直接写出线段$BE$长度的最小值。
(3)在射线$AC上另有一动点P$,且$AP = CD$,连接$BP$,取线段$BE的中点F$,当$AD + BP$的值最小时,求出此时线段$AF$的长度。

答案：
解：
(1)如图1，当点$D$在点$G$的右边时，过点$A$作$AG \perp BC$于点$G$。
$\because \angle ABC = \angle ACB = 30^{\circ}$，$AB = 2$，$\therefore AB = AC = 2$，$BG = CG$，$\therefore AG = \frac{1}{2}AB = 1$，$\therefore BG = CG = \sqrt{2^{2}-1^{2}} = \sqrt{3}$，$\therefore BC = 2\sqrt{3}$。
$\because AD$与$BC$所夹的锐角为$45^{\circ}$，$\therefore \angle ADG = \angle GAD = 45^{\circ}$，$\therefore AG = DG = 1$，$\therefore CD = \sqrt{3} - 1$。
当点$D$在点$G$的左边时，如图2所示。
同理可得$CD = \sqrt{3} + 1$。
综上所述，$CD$的长为$\sqrt{3} - 1$或$\sqrt{3} + 1$。


(2)①如图3，以$AB$，$AC$为边分别作等边$\triangle ABI$和等边$\triangle ACH$，连接$IH$，则$\angle HAC = 60^{\circ}$。又$\angle BAC = 180^{\circ} - 2×30^{\circ} = 120^{\circ}$，$\therefore \angle BAC + \angle HAC = 180^{\circ}$，$\therefore B$，$A$，$H$三点共线。
当点$D$，$C$重合时，点$E$，$H$重合；当点$D$，$B$重合时，点$E$，$I$重合，$\therefore$线段$HI$的长度是点$E$的移动路径的长度。$\because \triangle ABI$，$\triangle ACH$是等边三角形，$\therefore AB = BI = AI = 2$，$AC = AH = 2$，$\angle BAI = \angle BIA = 60^{\circ}$，$\therefore AI = AH$，$\therefore \angle IAH = 180^{\circ} - \angle BAI = 120^{\circ}$，$\therefore \angle AIH = \angle AHI = 30^{\circ}$，$\therefore \angle BIH = \angle BIA + \angle AIH = 90^{\circ}$，$\therefore HI = \sqrt{4^{2}-2^{2}} = 2\sqrt{3}$，$\therefore$点$E$的移动路径的长度为$2\sqrt{3}$。
②当$BE \perp HI$时，线段$BE$的长度最小，线段$BE$长度的最小值是$2$。

(3)如图4，过点$A$作$AG \perp BC$于点$G$，以$AC$为边作等边$\triangle ACH$，作$\angle BCM = 120^{\circ}$，$CM = CA$，连接$DM$，$EH$。又$\because AB = AC$，$\angle BAC = 120^{\circ}$，$\therefore CM = AB$，$\angle BAC = \angle DCM = 120^{\circ}$。
又$\because CD = AP$，$\therefore \triangle DCM \cong \triangle PAB(SAS)$，$\therefore DM = PB$，$\therefore AD + BP = AD + DM \geq AM$。
当$A$，$D$，$M$三点共线时，$AD + BP$的值最小，此时$\angle ACM = 120^{\circ} + 30^{\circ} = 150^{\circ}$，$\therefore \angle CAM = \angle CMA = 15^{\circ}$，$\therefore \angle CDM = 180^{\circ} - 120^{\circ} - 15^{\circ} = 45^{\circ} = \angle ADG$。
结合
(1)可得$AG = DG = 1$，$BG = CG = \sqrt{3}$，$\therefore CD = \sqrt{3} - 1$。
延长$FA$至点$Q$，使$AQ = AF$，连接$QE$，$QH$。
易证$\triangle ACD \cong \triangle AHE$，$\therefore EH = CD = \sqrt{3} - 1$。
$\because AB = AH$，$\angle FAB = \angle QAH$，$AF = AQ$，$\therefore \triangle ABF \cong \triangle AHQ(SAS)$，$\therefore BF = HQ$，$\angle ABF = \angle AHQ$，$\therefore QH // BE$，$\therefore \angle FEQ = \angle HQE$。
$\because F$为$BE$的中点，$\therefore BF = EF$，$\therefore EF = HQ$。
又$\because QE = EQ$，$\therefore \triangle FQE \cong \triangle HEQ(SAS)$，
$\therefore QF = HE = \sqrt{3} - 1$，$\therefore AF = \frac{1}{2}QF = \frac{\sqrt{3} - 1}{2}$。