《2025年名校题库九年级数学全一册北师大版》

2.(金牛区二诊)在$Rt\triangle ABC$中,$∠ACB= 90^{\circ },AB= 5,AC= 3$.
(1)如图1,点D在BC边上,$DE⊥AB$,垂足为E,已知$CD= DE$,求BE的长.
(2)将(1)中的$Rt\triangle BDE$绕点B顺时针旋转,连接CE,交直线AB于点G,在CE上方作$∠FCE= ∠ABC,∠FCE$的边与AB的交点为F.
①如图2,当点D落在CE上时,求BG的长;
②如图3,连接AD,延长CF交AD于点M,在$Rt\triangle BDE$旋转的过程中,若点M落在BE的垂直平分线上,求此时AM的长.

答案：
解:
(1)
∵∠ACB = 90°,DE⊥AB,
∴∠BED = 90° = ∠ACB.又
∵∠B = ∠B,
∴△ABC∽△DBE,
∴$\frac{DE}{AC}$=$\frac{BE}{BC}$=$\frac{BD}{AB}$.
由CD = DE,不妨设CD = x,则DE = x.
∵∠ACB = 90°,AB = 5,AC = 3,
∴$BC = \sqrt{AB^{2}-AC^{2}} = 4$,
∴BD = 4 - x,
∴$\frac{x}{3}=\frac{BE}{4}=\frac{4 - x}{5}$,解得$x=\frac{3}{2}$,
∴BE = 2.
(2)①如图1,延长BD交CF于点H.

∵∠FCE = ∠ABC,∠DBE = ∠ABC,
∴∠FCE = ∠DBE.
∵∠HDC = ∠BDE,
∴∠BHC = ∠CEB = 90°.
∵∠ABC = ∠DBE,
∴∠HBC = ∠EBG,
∴△HBC∽△EBG,
∴$\frac{BE}{BH}$=$\frac{BG}{BC}$.
在Rt△BEC中,BC = 4,BE = 2,
∴$CE = \sqrt{BC^{2}-BE^{2}} = 2\sqrt{3}$,
∴CD = CE - DE = $2\sqrt{3}-\frac{3}{2}$.
易证△CDH∽△BDE,
∴$\frac{DH}{CD}$=$\frac{DE}{BD}$=$\frac{3}{5}$,
∴$DH=\frac{3}{5}CD=\frac{6\sqrt{3}}{5}-\frac{9}{10}$,
∴$BH = BD + DH=\frac{6\sqrt{3}+8}{5}$,
∴$\frac{2}{\frac{6\sqrt{3}+8}{5}}=\frac{BG}{4}$,
∴$BG=\frac{60\sqrt{3}-80}{11}$.
②如图2,过点E作CE的垂线,与CM的延长线交于点N,连接ND,NB,MB,ME,AN,CD.
∵∠FCE = ∠ABC = ∠DBE,∠CEN = ∠BED = 90°,
∴△CEN∽△BED,
∴$\frac{NE}{DE} = \frac{CE}{BE}$,
∴$\frac{NE}{CE}=\frac{DE}{BE}$.
易得∠DEN = ∠CEB,
∴△DEN∽△BEC,
∴∠DNE = ∠BCE,$\frac{ND}{BC}$=$\frac{DE}{BE}$.
又
∵$\frac{DE}{BE}$=$\frac{AC}{BC}$,
∴$\frac{ND}{BC}$=$\frac{AC}{BC}$,即ND = AC.
∵∠DNE = ∠BCE,NE⊥CE,
∴ND⊥BC.
又
∵AC⊥BC,
∴ND//AC.
∵ND//AC,ND = AC,
∴四边形ACDN是平行四边形,
∴MC = MN.
在Rt△CEN中,EM是斜边CN的中线,
∴ME = MC = MN.
若点M在BE的垂直平分线上,则MB = ME,
∴MB = MC = MN,
∴△CBN为直角三角形,即NB⊥BC.
又ND⊥BC,
∴若点M在BE的垂直平分线上时,N,D,B三点共线.
如图2,当点B在ND的延长线上时,$AD = \sqrt{4^{2}+(3-\frac{5}{2})^{2}}=\frac{\sqrt{65}}{2}$,
∴$AM=\frac{1}{2}AD=\frac{\sqrt{65}}{4}$.

如图3,同理可得,当点B在线段ND上时,$AD=\sqrt{4^{2}+(3+\frac{5}{2})^{2}}=\frac{\sqrt{185}}{2}$,
∴$AM=\frac{\sqrt{185}}{4}$.
综上所述,当点M落在BE的垂直平分线上时,AM的长为$\frac{\sqrt{65}}{4}$或$\frac{\sqrt{185}}{4}$.