《2025年天府前沿九年级数学上册北师大版》

15. (2025·编写)探究式学习是新课程倡导的重要学习方式,某兴趣小组拟做以下探究. 在$Rt\triangle ABC$中,$\angle C= 90^{\circ}$,$AC= BC$,$D是AB$边上一点,且$\frac{AD}{BD}= \frac{1}{n}$($n$为正整数),$E是AC$边上的动点,过点$D作DE的垂线交直线BC于点F$.
【初步感知】(1)如图1,当$n= 1$时,兴趣小组探究得出结论：$AE+BF= \frac{\sqrt{2}}{2}AB$,请写出证明过程.
【深入探究】(2)①如图2,当$n= 2$,且点$F在线段BC$上时,试探究线段$AE$,$BF$,$AB$之间的数量关系,请写出结论并证明；
②请通过类比、归纳、猜想,探究出线段$AE$,$BF$,$AB$之间数量关系的一般结论.(直接写出结论,不必证明)

答案：

(1)如图1, 连接$CD$.
$\because ∠ACB = 90^{\circ}, AC = BC, AD = DB,$
$\therefore AB = \sqrt{2}AC, ∠A = ∠B = ∠ACD = 45^{\circ}, AD = CD = BD, CD \perp AB.$
$\because ED \perp FD, \therefore ∠EDF = ∠CDB = 90^{\circ},$
$\therefore ∠CDE = ∠BDF,$
$\therefore △CDE \cong △BDF(ASA), \therefore CE = BF,$
$\therefore AE + BF = AE + CE = AC = \frac{\sqrt{2}}{2}AB.$

(2)①$AE + \frac{1}{2}BF = \frac{\sqrt{2}}{3}AB$, 理由如下:
如图2, 过点$D$作$DN \perp AC$于点$N$,$DH \perp BC$于点$H$.
$\because ∠ACB = 90^{\circ}, AC = BC, \therefore ∠A = ∠B = 45^{\circ}.$
$\because DN \perp AC, DH \perp BC,$
$\therefore △ADN$和$△BDH$都是等腰直角三角形,
$\therefore AN = DN, DH = BH, AD = \sqrt{2}AN, BD = \sqrt{2}BH,$
$∠A = ∠B = 45^{\circ} = ∠ADN = ∠BDH,$
$\therefore △ADN \sim △BDH, \therefore \frac{AD}{DB} = \frac{AN}{BH} = \frac{1}{2}.$
设$AN = DN = x$, 则$BH = DH = 2x,$
$\therefore AD = \sqrt{2}x, BD = 2\sqrt{2}x, \therefore AB = 3\sqrt{2}x.$
$\because DN \perp AC, DH \perp BC, ∠ACB = 90^{\circ},$
$\therefore$ 四边形$DHCN$是矩形,
$\therefore ∠NDH = 90^{\circ} = ∠EDF, \therefore ∠EDN = ∠FDH.$
又$\because ∠END = ∠FHD, \therefore △EDN \sim △FDH,$
$\therefore \frac{EN}{FH} = \frac{DN}{DH} = \frac{1}{2}, \therefore FH = 2NE,$
$\therefore AE + \frac{1}{2}BF = x + NE + \frac{1}{2}(2x - FH) = 2x = \frac{\sqrt{2}}{3}AB.$

②如图3, 当点$F$在射线$BC$上时, 过点$D$作$DN \perp AC$于点$N$,$DH \perp BC$于点$H$.
$\because ∠ACB = 90^{\circ}, AC = BC, \therefore ∠A = ∠B = 45^{\circ}.$
$\because DN \perp AC, DH \perp BC,$
$\therefore △ADN$和$△BDH$都是等腰直角三角形,
$\therefore AN = DN, DH = BH, AD = \sqrt{2}AN, BD = \sqrt{2}BH,$
$∠A = ∠B = 45^{\circ} = ∠ADN = ∠BDH,$
$\therefore △ADN \sim △BDH, \therefore \frac{AD}{DB} = \frac{AN}{BH} = \frac{1}{n}.$
设$AN = DN = x$, 则$BH = DH = nx,$
$\therefore AD = \sqrt{2}x, BD = \sqrt{2}nx, \therefore AB = \sqrt{2}(n + 1)x.$
$\because DN \perp AC, DH \perp BC, ∠ACB = 90^{\circ},$
$\therefore$ 四边形$DHCN$是矩形,
$\therefore ∠NDH = 90^{\circ} = ∠EDF, \therefore ∠EDN = ∠FDH.$
又$\because ∠END = ∠FHD, \therefore △EDN \sim △FDH,$
$\therefore \frac{EN}{FH} = \frac{DN}{DH} = \frac{1}{n}, \therefore FH = nNE,$
$\therefore AE + \frac{1}{n}BF = x - NE + \frac{1}{n}(nx + FH) = 2x = \frac{\sqrt{2}}{n + 1}AB;$

当点$F$在$CB$的延长线上时, 如图4所示.
$\because ∠C = 90^{\circ}, AC = BC,$
$\therefore ∠A = ∠ABC = 45^{\circ}.$
$\because DN \perp AC, DH \perp BC,$
$\therefore △ADN$和$△BDH$都是等腰直角三角形,
$\therefore AN = DN, DH = BH, AD = \sqrt{2}AN, BD = \sqrt{2}BH,$
$∠A = ∠ABC = 45^{\circ} = ∠ADN = ∠BDH,$
$\therefore △ADN \sim △BDH, \therefore \frac{AD}{DB} = \frac{AN}{BH} = \frac{1}{n}.$
设$AN = DN = x$, 则$BH = DH = nx,$
$\therefore AD = \sqrt{2}x, BD = \sqrt{2}nx, \therefore AB = \sqrt{2}(n + 1)x.$
$\because DN \perp AC, DH \perp BC, ∠ACB = 90^{\circ},$
$\therefore$ 四边形$DHCN$是矩形,$\therefore ∠NDH = 90^{\circ} = ∠EDF,$
$\therefore ∠EDN = ∠FDH.$
又$\because ∠END = ∠FHD,$
$\therefore △EDN \sim △FDH, \therefore \frac{EN}{FH} = \frac{DN}{DH} = \frac{1}{n},$
$\therefore FH = nNE,$
$\therefore AE - \frac{1}{n}BF = x + NE - \frac{1}{n}(FH - nx) = 2x = \frac{\sqrt{2}}{n + 1}AB.$
综上所述, 当点$F$在射线$BC$上时,$AE + \frac{1}{n}BF = \frac{\sqrt{2}}{n + 1}AB$; 当点$F$在$CB$延长线上时,$AE - \frac{1}{n}BF = \frac{\sqrt{2}}{n + 1}AB.$