《2025年天府前沿九年级数学上册北师大版》

14. (2023·通州)如图,在△ABC中,∠ABC= 90°,AB= 3BC,AD= 3,BD= 6,连接CD,求CD长的最大值.

答案：
【解】如图，在$AD$的下方作$\mathrm{Rt}\triangle ADT$，使得$\angle ADT = 90^{\circ}$，$DT = 1$，连接$CT$，则$AT = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$。

$\because$在$\triangle ABC$中，$\angle ABC = 90^{\circ}$，$AB = 3BC$，$\therefore \frac{AD}{DT} = \frac{AB}{BC} = 3$，$\therefore \frac{AD}{AB} = \frac{DT}{BC}$。$\because \angle ADT = \angle ABC = 90^{\circ}$，$\therefore \triangle ADT \backsim \triangle ABC$，$\therefore \angle DAT = \angle BAC$，$\frac{AD}{AB} = \frac{AT}{AC}$，$\therefore \angle DAB = \angle TAC$，$\therefore \triangle DAB \backsim \triangle TAC$，$\therefore \frac{DB}{TC} = \frac{AD}{AT} = \frac{3}{\sqrt{10}}$，$\therefore TC = 2\sqrt{10}$。$\because CD \leq DT + CT$，$\therefore CD \leq 1 + 2\sqrt{10}$，$\therefore CD$长的最大值为$1 + 2\sqrt{10}$。

15. (2025·金牛)【问题探究】
(1)如图1,在矩形ABCD中,E,F分别是边DC,BC上的点,连接AE,DF,且AE⊥DF于点G.若AB= 10,BC= 24,求DF/AE的值.
【初步运用】

(2)如图2,在△ABC中,∠BAC= 90°,AB/AC = 5/12,D为AC的中点,连接BD,过点A作AE⊥BD于点E,交BC于点F,求AF/BD的值.
【灵活运用】
(3)如图3,在四边形ABCD中,∠BAD= 90°,AB/AD = 5/12,AB= BC,AD= CD,点E,F分别在边AB,AD上,且DE⊥CF,垂足为G,则CF/DE = ______.

答案：
【解】
(1)
∵四边形$ABCD$是矩形，$\therefore \angle C = \angle ADE = 90^{\circ}$，$AB = CD = 10$，$BC = AD = 24$，$\therefore \angle ADG + \angle FDC = 90^{\circ}$。$\because AE \perp DF$，$\therefore \angle AGD = 90^{\circ}$，$\therefore \angle DAG + \angle ADG = 90^{\circ}$，$\therefore \angle DAG = \angle FDC$，$\therefore \triangle DCF \backsim \triangle ADE$，$\therefore \frac{DF}{AE} = \frac{DC}{AD} = \frac{10}{24} = \frac{5}{12}$。
(2)如图，过点$B$作$CB$的垂线，过点$D$作$BC$的垂线，垂足为$K$，过点$A$作$BC$的平行线，分别交两条垂线于点$G$，$H$，则四边形$GBKH$为矩形。

$\because D$为$AC$的中点，$\therefore AD = CD$。又$\because \angle AHD = \angle CKD = 90^{\circ}$，$\angle ADH = \angle CDK$，$\therefore \triangle ADH \cong \triangle CDK(AAS)$，$\therefore DH = DK$。$\because \angle BAD = 90^{\circ}$，$\therefore \angle GAB + \angle HAD = 90^{\circ}$。又$\because \angle GAB + \angle GBA = 90^{\circ}$，$\therefore \angle HAD = \angle GBA$。又$\because \angle AHD = \angle G = 90^{\circ}$，$\therefore \triangle AHD \backsim \triangle BGA$，$\therefore \frac{DH}{AG} = \frac{AH}{BG} = \frac{AD}{AB}$。$\because \frac{AB}{AC} = \frac{5}{12}$，$AD = CD$，$\therefore \frac{AD}{AB} = \frac{6}{5}$。设$DH = 6y$，则$BG = 12y$，$\therefore \frac{6y}{AG} = \frac{AH}{12y} = \frac{6}{5}$，$\therefore AG = 5y$，$AH = \frac{72}{5}y$，$\therefore \frac{BG}{GH} = \frac{12y}{5y + \frac{72}{5}y} = \frac{60}{97}$。由
(1)知，$\frac{AF}{BD} = \frac{BG}{GH}$，$\therefore \frac{AF}{BD}$的值为$\frac{60}{97}$。
(3)如图，过点$C$作$CN \perp AD$于点$N$，$CM \perp AB$交$AB$的延长线于点$M$，连接$BD$。

$\because \angle BAD = 90^{\circ}$，即$AB \perp AD$，$\therefore \angle A = \angle M = \angle CNA = 90^{\circ}$，$\therefore$四边形$AMCN$是矩形，$\therefore AM = CN$，$AN = CM$。在$\triangle BAD$和$\triangle BCD$中，$\begin{cases} AD = CD \\ AB = BC \\ BD = BD \end{cases}$，$\therefore \triangle BAD \cong \triangle BCD(SSS)$，$\therefore \angle BCD = \angle A = 90^{\circ}$，$\therefore \angle ABC + \angle ADC = 180^{\circ}$。$\because \angle ABC + \angle CBM = 180^{\circ}$，$\therefore \angle MBC = \angle ADC$。又$\because \angle CND = \angle M = 90^{\circ}$，$\therefore \triangle BCM \backsim \triangle DCN$，$\therefore \frac{BM}{DN} = \frac{BC}{CD} = \frac{AB}{AD} = \frac{5}{12}$。设$BM = 5y$，则$DN = 12y$。设$AB = BC = 5x$，则$AD = CD = 12x$，$\therefore CN = AM = AB + BM = 5x + 5y$。在$\mathrm{Rt}\triangle CND$中，由勾股定理，得$DN^{2} + CN^{2} = CD^{2}$，$\therefore (12y)^{2} + (5x + 5y)^{2} = (12x)^{2}$，解得$119x = 169y (x = -y$舍去)。$\therefore \frac{CF}{DE} = \frac{CN}{AD} = \frac{5x + 5y}{12x} = \frac{120}{169}$。
故答案为$\frac{120}{169}$。