《2025年名校题库八年级数学上册北师大版》

9.(嘉祥)如图,$\triangle ACB和\triangle DCE$都是直角三角形,$∠ACB= ∠DCE= 90^{\circ },CB= CD,∠B= 45^{\circ }$.
(1)如图1,$BC= 2,DE= 3$,求AE的长度.
(2)如图2,延长BA交DE于点F,取AB的中点G,连接AD,DG.若$AG= AF= 2$,求$\triangle AEF$的面积.
(3)如图3,在(2)的条件下,连接EB,过点C作$CM⊥DE$于点M,延长MC交EB的延长线于点N,直接写出$\frac {BN}{CN}$的值.

答案：
解:
(1)
∵BC = 2,∠B = 45°,∠ACB = 90°,
∴△ABC是等腰直角三角形,
∴AC = BC = 2.在Rt△CDE中,
∵CB = CD,
∴CD = 2.
∵DE = 3,
∴CE = $\sqrt{DE^{2} - CD^{2}}$ = $\sqrt{3^{2} - 2^{2}}$ = $\sqrt{5}$,
∴AE = CE - AC = $\sqrt{5} - 2$.
(2)如图1,过点D作DH⊥BD,交BF的延长线于点H,过点F作FK⊥AE于点K,则∠BDH = 90°.

∵∠ACB + ∠DCE = 180°,
∴B,C,D三点在同一条直线上.
∵∠B = 45°,
∴△BDH是等腰直角三角形,
∴DH = BD,∠H = ∠B = 45°.
∵G是AB的中点,AG = AF = 2,
∴AB = 4.
∵∠ACB = ∠DCE = 90°,∠B = 45°,
∴△ABC是等腰直角三角形,
∴AC = BC = 2$\sqrt{2}$,∠BAC = ∠B = 45°.
∵BC = CD,
∴AC = CD,
∴△ACD是等腰直角三角形,
∴∠CAD = ∠ADC = 45°,
∴∠BAD = ∠BAC + ∠CAD = 90°.在△ADF和△ADG中,$\begin{cases} AF = AG, \\ \angle DAF = \angle DAG, \\ AD = AD, \end{cases}$
∴△ADF ≌ △ADG(SAS),
∴∠ADF = ∠ADG,DF = DG.
∵∠ADH = ∠BDH - ∠ADB = 90° - 45° = 45°,
∴∠ADH = ∠ADB,
∴∠ADH - ∠ADF = ∠ADB - ∠ADG,即∠FDH = ∠GDB.在△DFH和△DGB中,$\begin{cases} DH = DB, \\ \angle FDH = \angle GDB, \\ DF = DG, \end{cases}$
∴△DFH ≌ △DGB(SAS),
∴FH = BG = 2,
∴AF = FH.
∵CB = CD,
∴BD = 4$\sqrt{2}$,
∴DH = DB = 4$\sqrt{2}$.
∵∠ACB = ∠BDH = 90°,
∴CE // DH,
∴∠EAF = ∠H.在△AEF和△HDF中,$\begin{cases} \angle EAF = \angle H, \\ AF = HF, \\ \angle AFE = \angle HFD, \end{cases}$
∴△AEF ≌ △HDF(ASA),
∴AE = DH = 4$\sqrt{2}$.
∵∠FAK = ∠BAC = 45°,∠AKF = 90°,
∴△AFK是等腰直角三角形,
∴FK = $\frac{\sqrt{2}}{2}AF$ = $\sqrt{2}$,
∴$S_{\triangle AEF} = \frac{1}{2}AE \cdot FK = \frac{1}{2}×4\sqrt{2}×\sqrt{2} = 4$.
(3)$\frac{1}{3}$ [解析]如图2,过点C作CP⊥BE于点P,过点B作BT⊥MN于点T.

由
(2)知AC = BC = CD = 2$\sqrt{2}$,AE = 4$\sqrt{2}$,
∴CE = AC + AE = 2$\sqrt{2} + 4\sqrt{2} = 6\sqrt{2}$.在Rt△CDE中,DE = $\sqrt{CD^{2} + CE^{2}}$ = $\sqrt{(2\sqrt{2})^{2} + (6\sqrt{2})^{2}}$ = 4$\sqrt{5}$.
∵∠ACB = 90°,CB = CD,
∴CE垂直平分线段BD,
∴BE = DE.
∵CM⊥DE于点M,
∴$S_{\triangle CDE} = \frac{1}{2}CD \cdot CE = \frac{1}{2}DE \cdot CM$,
∴CM = $\frac{CD \cdot CE}{DE}$ = $\frac{2\sqrt{2} × 6\sqrt{2}}{4\sqrt{5}}$ = $\frac{6\sqrt{5}}{5}$,
∴CP = $\frac{6\sqrt{5}}{5}$.在Rt△CDM中,DM = $\sqrt{CD^{2} - CM^{2}}$ = $\sqrt{(2\sqrt{2})^{2} - (\frac{6\sqrt{5}}{5})^{2}}$ = $\frac{2\sqrt{5}}{5}$.在△CBT和△CDM中,$\begin{cases} \angle CTB = \angle CMD = 90^{\circ}, \\ \angle BCT = \angle DCM, \\ BC = DC, \end{cases}$
∴△CBT ≌ △CDM(AAS),
∴BT = DM = $\frac{2\sqrt{5}}{5}$.
∵$S_{\triangle BCN} = \frac{1}{2}BN \cdot CP = \frac{1}{2}CN \cdot BT$,
∴$\frac{6\sqrt{5}}{5}BN = \frac{2\sqrt{5}}{5}CN$,即3BN = CN,
∴$\frac{BN}{CN} = \frac{1}{3}$.

1. (实外)如图, 在$\triangle ABC$中,$AB = AC$,$P为BC$边上任意一点, 请用学过的知识说明:$AB^{2}-AP^{2}= PB\cdot PC$.

答案：
解:如图,过点A作$AE⊥BC$,垂足为E,则$AB^{2}=AE^{2}+BE^{2},AP^{2}=PE^{2}+AE^{2}$.又$\because AB=AC,\therefore BE=EC,\therefore AB^{2}-AP^{2}=(AE^{2}+BE^{2})-(PE^{2}+AE^{2})=BE^{2}-PE^{2}=(BE+PE)(BE-PE)=(EC+PE)\cdot PB=PC\cdot PB$.