《2025年名校题库九年级数学全一册北师大版》

2.(天府新区一诊)阅读下面的材料:
天府新区某学校数学兴趣活动小组在一次活动中对一个数学问题进行了如下探究:
(1)问题发现:如图1,在$\triangle ABC$中,$AB= AC$,点D在BC边上,$∠DAB= ∠ABD,BE⊥AD$,垂足为E.小明经探究发现,过点A作$AF⊥BC$,垂足为F,可得到$BC= 2AE$,请你写出证明过程.
(2)变式探究:如图2,在$\triangle ABC$中,$AB= AC= 6,∠BAC= 90^{\circ }$,D为BC的中点,E为DC的中点,点F在AC的延长线上,且$∠CDF= ∠EAC$,求CF的长.
(3)解决问题:如图3,在$\triangle ABC$中,$AB= AC,∠BAC= 120^{\circ }$,点D,E分别在AB,AC边上,且$AD= kDB$(其中$0＜k＜\frac {\sqrt {3}}{3}),∠AED= ∠BCD$,求$\frac {AE}{EC}$的值(用含k的式子表示).

答案：

(1)证明:如图1,过点A作AF⊥BC于点F.
∵BE⊥AD,
∴∠AFB = ∠BEA = 90°.
又
∵∠DAB = ∠ABD,AB = AB,
∴△ABF≌△BAE(AAS),
∴BF = AE.
∵AB = AC,AF⊥BC,
∴$BF=\frac{1}{2}BC$,
∴BC = 2AE.

(2)解:如图2,连接AD,过点C作CG⊥DF于点G,过点E作EK⊥AC于点K.

在Rt△ABC中,
∵∠BAC = 90°,AB = AC = 6,D是BC的中点,
∴AD = CD = BD = 3$\sqrt{2}$,∠ACB = 45°.
又
∵E是DC的中点,
∴$DE=\frac{1}{2}CD=\frac{3\sqrt{2}}{2}$.
∵△CEK是等腰直角三角形,
∴EK = CK=$\frac{3}{2}$,
∴$AK = AC - CK = 6 - \frac{3}{2}=\frac{9}{2}$,
∴$AE=\sqrt{AK^{2}+EK^{2}}=\frac{3\sqrt{10}}{2}$.
∵∠EAK = ∠CDG,∠AKE = ∠CGD = 90°,
∴△AKE∽△DGC,
∴$\frac{AE}{CD}$=$\frac{EK}{CG}$,
∴$\frac{\frac{3\sqrt{10}}{2}}{3\sqrt{2}}=\frac{\frac{3}{2}}{CG}$,
∴$CG=\frac{3\sqrt{5}}{5}$.
∵AB = AC,∠BAC = 90°,D为BC的中点,
∴∠ADC = 90°,∠ACB = ∠DAC = 45°,
∴∠F + ∠CDF = ∠ACB = 45°.
∵∠CDF = ∠EAC,
∴∠F + ∠EAC = 45°.
又
∵∠DAE + ∠EAC = 45°,
∴∠F = ∠DAE,
∴△ADE∽△FGC,
∴$\frac{CG}{FG}$=$\frac{DE}{AD}$=$\frac{1}{2}$,
∴$FG=\frac{6\sqrt{5}}{5}$,
∴$CF=\sqrt{CG^{2}+FG^{2}} = 3$.
(3)解:如图3,过点D作DG⊥BC于点G,设DG = a.
∵AB = AC,∠A = 120°,
∴∠B = ∠ACB = 30°.

在Rt△BGD中,
∵∠B = 30°,
∴BD = 2a,BG=$\sqrt{3}a$.
∵AD = kDB,
∴AD = 2ka,
∴AB = BD + AD = 2a + 2ka = 2a(k + 1).
如图3,过点A作AH⊥BC于点H.
在Rt△ABH中,
∵∠B = 30°,
∴$BH=\sqrt{3}a(k + 1)$.
∵AB = AC,AH⊥BC,
∴BC = 2BH = $2\sqrt{3}a(k + 1)$,
∴$CG = BC - BG=\sqrt{3}a(2k + 1)$.
如图3,过点D作DN⊥AC交CA的延长线于点N.
∵∠BAC = 120°,
∴∠DAN = 60°,
∴∠ADN = 30°,
∴AN = ka,DN=$\sqrt{3}ka$.
∵∠DGC = ∠AND = 90°,∠AED = ∠BCD,
∴△NDE∽△GDC,
∴$\frac{DN}{DG}$=$\frac{NE}{CG}$,
∴$\frac{\sqrt{3}ka}{a}=\frac{NE}{\sqrt{3}a(2k + 1)}$,
∴NE = 3ak(2k + 1).
又
∵AN = ka,
∴AE = NE - AN = 2ak(3k + 1),
∴EC = AC - AE = AB - AE = 2a(k + 1)-2ak(3k + 1)=2a(1 - 3k^{2}),
∴$\frac{AE}{EC}$=$\frac{2ak(3k + 1)}{2a(1 - 3k^{2})}$=$\frac{3k^{2}+k}{1 - 3k^{2}}$.