《2025年名校题库九年级数学全一册北师大版》

4.(嘉祥)问题背景:如图1,在$\triangle ABC和\triangle AED$中,若$\frac {AB}{AC}= \frac {AE}{AD}= 1,∠BAC= ∠EAD= 60^{\circ }$,易知$BE= CD$,BE与CD的夹角为$60^{\circ }$.
问题提出:(1)如图2,若$\frac {AB}{AC}= \frac {AE}{AD}= \frac {1}{2},∠BAC= ∠EAD= α$,则BE与CD的数量关系为______,它们的夹角为______.
尝试运用:(2)如图3,若$\frac {AB}{AC}= \frac {AE}{AD}= \frac {1}{3},∠BAC= ∠EAD= 90^{\circ }$,E为线段BC上一点(不与点B重合),AC与ED相交于点M,连接CD,若$AB= 2,\frac {DC}{CE}= \frac {1}{2}$,求DE的长度.
拓展创新:(3)如图4,在$Rt\triangle ABC$中,$∠BAC= 90^{\circ },AB= 5,AC= 12$,D,E分别为BC,AC上一点,$AD= \frac {60}{13}\sqrt {2},∠EDA= ∠B$,求$\frac {DE}{AE}$的值.

答案：

(1)$2BE = CD$ $\alpha$ 【解析】
∵∠BAC = ∠EAD = α,
∴∠BAC - ∠EAC = ∠EAD - ∠EAC,
∴∠BAE = ∠CAD.又
∵$\frac{AB}{AC}$=$\frac{AE}{AD}$ = $\frac{1}{2}$,
∴△ABE∽△ACD,
∴$\frac{AB}{AC}$=$\frac{BE}{CD}$=$\frac{1}{2}$,∠ABE = ∠ACD,
∴2BE = CD.如图1,延长BE,CD交于点F,BF交AC于点O.
∵∠AOB = ∠COF,
∴∠F = ∠BAC = α.
(2)解:
∵$\frac{AB}{AC}$=$\frac{AE}{AD}$=$\frac{1}{3}$,∠BAC = ∠EAD = 90°,
∴△BAE∽△CAD,
∴$\frac{BE}{CD}$=$\frac{1}{3}$,即CD = 3BE.
∵∠BAE = ∠CAD,
∴∠BCD = 90°.
∵AB = 2,
∴AC = 6,
∴由勾股定理,得$BC = 2\sqrt{10}$.
设BE = x,则CE = 2$\sqrt{10}-x$,CD = 3x.
∵$\frac{CD}{CE}$=$\frac{1}{2}$,
∴CE = 6x,
∴$2\sqrt{10}-x = 6x$,解得$x=\frac{2\sqrt{10}}{7}$,
∴$CD=\frac{6\sqrt{10}}{7}$,$CE=\frac{12\sqrt{10}}{7}$.
在Rt△CDE中,$DE=\sqrt{CD^{2}+CE^{2}}=\frac{30\sqrt{2}}{7}$.

(3)解:如图2,过点A作AF⊥BD于点F,过点D作DG⊥AB于点G,过点E作EH⊥AD于点H,
∴∠AFB = ∠BAC = 90°.
在Rt△ABC中,
∵AB = 5,AC = 12,
∴BC = 13.
∵∠AFB = ∠BAC,∠B = ∠B,
∴△ABF∽△CBA,
∴$\frac{AB}{BC}$=$\frac{BF}{BA}$,
∴$\frac{5}{13}$=$\frac{BF}{5}$,
∴$BF=\frac{25}{13}$,
∴$AF=\sqrt{AB^{2}-BF^{2}}=\sqrt{5^{2}-(\frac{25}{13})^{2}}=\frac{60}{13}$,
∴$DF=\sqrt{AD^{2}-AF^{2}}=\frac{60}{13}$,
∴BD = BF + DF = $\frac{85}{13}$.
∵∠BGD = ∠BAC = 90°,
∴DG//AC,
∴△BDG∽△BCA,
∴$\frac{BG}{BA}$=$\frac{BD}{BC}$,即$\frac{BG}{5}$=$\frac{\frac{85}{13}}{13}$,
∴$BG=\frac{425}{169}$,
∴$AG = AB - BG=\frac{420}{169}$.
∵∠BAD + ∠DAC = ∠AEH + ∠CAD = 90°,
∴∠BAD = ∠AEH.
又
∵∠AHE = ∠AGD = 90°,
∴△AEH∽△DAG,
∴$\frac{AE}{AD}$=$\frac{EH}{AG}$.
设DH = x.
∵∠EDA = ∠B,∠DHE = ∠BAC = 90°,
∴△DHE∽△BAC,
∴$\frac{DE}{BC}$=$\frac{EH}{AC}$=$\frac{DH}{AB}$,即$\frac{DE}{13}$=$\frac{EH}{12}$=$\frac{x}{5}$,
∴$DE=\frac{13}{5}x$,$EH=\frac{12}{5}x$,
∴$AE=\frac{AD\cdot EH}{AG}=\frac{\frac{60\sqrt{2}}{13}\cdot \frac{12x}{5}}{\frac{420}{169}}=\frac{156\sqrt{2}}{35}x$,
∴$\frac{DE}{AE}$=$\frac{\frac{13}{5}x}{\frac{156\sqrt{2}}{35}x}$=$\frac{7\sqrt{2}}{24}$.