《2026年启东中学作业本九年级数学下册苏科版》

1. 如图，在$Rt△ABC$中，$∠ABC = 90^{\circ}$，$AB = 4$，$BC = 2$，$BD$是边$AC$上的高.点$E$，$F$分别在边$AB$，$BC$上(不与端点重合)，且$DE⊥DF$.设$AE = x$，四边形$DEBF$的面积为$y$，则$y$关于$x$的函数图像为(A)

答案： 1. A

2. 如图，在$△ABC$中，$∠ACB = 90^{\circ}$，$CD⊥AB$于点$D$，$E$，$F$分别为$AC$，$BC$上的点，且$DE⊥DF$，$\frac{BD}{AD} = \frac{1}{4}$.
求：（1）$\frac{BC}{AC}$的值；
（2）$\frac{DF}{DE}$的值.

答案： 2. 解:
(1)
∵CD⊥AB,
∴∠ADC = ∠CDB = 90°.
∵∠ACB = 90°,
∴∠ACD + ∠BCD = 90°.
又
∵∠BCD + ∠B = 90°,
∴∠ACD = ∠B,
∴△ACD∽△CBD,
∴$\frac{CD}{BD}$ = $\frac{AD}{CD}$,即CD² = AD·BD.
∵$\frac{BD}{AD}$ = $\frac{1}{4}$,
∴设BD = x,则AD = 4x,
∴CD² = 4x²,
∴CD = 2x.
∵△ACD∽△CBD,
∴$\frac{BC}{AC}$ = $\frac{BD}{CD}$ = $\frac{x}{2x}$ = $\frac{1}{2}$.
(2)
∵DE⊥DF,
∴∠EDF = 90°,
∴∠EDC + ∠CDF = 90°.
∵∠CDB = 90°,
∴∠CDF + ∠FDB = 90°,
∴∠EDC = ∠FDB.
由
(1)知∠ACD = ∠B,
∴△FDB∽△EDC,
∴$\frac{DF}{DE}$ = $\frac{BD}{CD}$ = $\frac{x}{2x}$ = $\frac{1}{2}$.

3. 如图，在$△ABC$中，$∠ACB = 90^{\circ}$，$CD⊥AB$，垂足为$D$，$E$是$BC$上一点，连接$AE$交$CD$于点$F$，且$AC^{2} = CD· AE$.
（1）求证：$∠CEA = ∠CAB$；
（2）作$CG⊥AE$，垂足为$G$，延长$CG$交$AB$于点$M$，求证：$AE· CM = AB· CF$.

答案：
3. 证明:
(1)
∵AC² = CD·AE,
∴$\frac{AE}{AC}$ = $\frac{AC}{CD}$.
设$\frac{AE}{AC}$ = $\frac{AC}{CD}$ = k,则AE = kAC,AC = kCD,
在Rt△ACE中,CE = $\sqrt{AE^{2} - AC^{2}}$ = $\sqrt{k^{2} - 1}$AC,
在Rt△ACD中,AD = $\sqrt{AC^{2} - CD^{2}}$ = $\sqrt{k^{2} - 1}$CD,
∴$\frac{CE}{AC}$ = $\sqrt{k^{2} - 1}$ = $\frac{AD}{CD}$.
又
∵∠ACE = ∠CDA = 90°,
∴△ACE∽△CDA,
∴∠AEC = ∠CAD,即∠CEA = ∠CAB.
(2) 补全图形如答图.

由
(1)知∠CEA = ∠CAB,又
∵∠ACE = ∠BCA = 90°,
∴△ACE∽△BCA,
∴$\frac{BC}{AC}$ = $\frac{AB}{AE}$,∠CAE = ∠B.
∵∠CAD = ∠BAC,∠ADC = ∠ACB = 90°,
∴△ACD∽△ABC,
∴$\frac{CD}{BC}$ = $\frac{AD}{AC}$,∠ACD = ∠B,
∴$\frac{CD}{AD}$ = $\frac{BC}{AC}$,∠CAE = ∠ACD,
∴$\frac{AB}{AE}$ = $\frac{CD}{AD}$,AF = CF.
∵CD⊥AB,CG⊥AE,
∴∠CDM = ∠ADF = ∠CGF = 90°.
又
∵∠AFD = ∠CFG,
∴∠DCM = ∠DAF,
∴△DCM∽△DAF,
∴$\frac{CM}{AF}$ = $\frac{CD}{AD}$,
∴$\frac{CM}{AF}$ = $\frac{AB}{AE}$,又
∵AF = CF,
∴$\frac{CM}{CF}$ = $\frac{AB}{AE}$,
∴AE·CM = AB·CF.