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1. 如图，在$Rt\triangle ABC$中，$\angle ABC = 90^{\circ}$，$AB = 4$，$BC = 2$，$BD$是边$AC$上的高. 点$E$，$F$分别在边$AB$，$BC$上(不与端点重合)，且$DE\perp DF$. 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 点拨:过点D作DG⊥AB于点G,作DH⊥BC于点H,如答图.
∵在Rt△ABC中,BD⊥AC,由勾股定理，得$AC = \sqrt{AB^{2} + BC^{2}} = 2\sqrt{5}$，则$BD = \frac{AB· BC}{AC} = \frac{4 × 2}{2\sqrt{5}} = \frac{4\sqrt{5}}{5}$.易证△ABC∽△ADB∽△BDC,
∴$\frac{BC}{AB} = \frac{BD}{AD} = \frac{CD}{BD} = \frac{2}{4} = \frac{1}{2}$，
∴$CD = \frac{1}{2}BD = \frac{2\sqrt{5}}{5}$，$AD = 2BD = \frac{8\sqrt{5}}{5}$.
在Rt△BDC中,
∵∠BDC = 90°，
∴$DH· BC = CD· BD$，
∴$DH = \frac{CD· BD}{BC} = \frac{\frac{2\sqrt{5}}{5} × \frac{4\sqrt{5}}{5}}{2} = \frac{4}{5}$.
在Rt△ADB中,
∵∠ADB = 90°，
∴$AD· DB = DG· AB$，
∴$DG = \frac{AD· DB}{AB} = \frac{\frac{8\sqrt{5}}{5} × \frac{4\sqrt{5}}{5}}{4} = \frac{8}{5}$.
∵∠ADB = ∠EDF = 90°，
∴∠ADB - ∠BDE = ∠EDF - ∠BDE，即∠ADE = ∠BDF.
又
∵∠DAE + ∠ABD = ∠DBF + ∠ABD = 90°，
∴∠DAE = ∠DBF，
∴△ADE∽△BDF，
∴$\frac{BF}{AE} = \frac{BD}{AD} = \frac{\frac{4\sqrt{5}}{5}}{\frac{8\sqrt{5}}{5}} = \frac{1}{2}$，则$BF = \frac{1}{2}AE = \frac{1}{2}x$，
∴$y = S_{四边形BEDF} = S_{\triangle BDE} + S_{\triangle BDF} = \frac{1}{2}(4 - x) × \frac{8}{5} + \frac{1}{2} × \frac{1}{2}x × \frac{4}{5} = -\frac{3}{5}x + \frac{16}{5}(0 ＜ x ＜ 4)$.
1.A 点拨:过点D作DG⊥AB于点G,作DH⊥BC于点H,如答图.
∵在Rt△ABC中,BD⊥AC,由勾股定理，得$AC = \sqrt{AB^{2} + BC^{2}} = 2\sqrt{5}$，则$BD = \frac{AB· BC}{AC} = \frac{4 × 2}{2\sqrt{5}} = \frac{4\sqrt{5}}{5}$.易证△ABC∽△ADB∽△BDC,
∴$\frac{BC}{AB} = \frac{BD}{AD} = \frac{CD}{BD} = \frac{2}{4} = \frac{1}{2}$，
∴$CD = \frac{1}{2}BD = \frac{2\sqrt{5}}{5}$，$AD = 2BD = \frac{8\sqrt{5}}{5}$.
在Rt△BDC中,
∵∠BDC = 90°，
∴$DH· BC = CD· BD$，
∴$DH = \frac{CD· BD}{BC} = \frac{\frac{2\sqrt{5}}{5} × \frac{4\sqrt{5}}{5}}{2} = \frac{4}{5}$.
在Rt△ADB中,
∵∠ADB = 90°，
∴$AD· DB = DG· AB$，
∴$DG = \frac{AD· DB}{AB} = \frac{\frac{8\sqrt{5}}{5} × \frac{4\sqrt{5}}{5}}{4} = \frac{8}{5}$.
∵∠ADB = ∠EDF = 90°，
∴∠ADB - ∠BDE = ∠EDF - ∠BDE，即∠ADE = ∠BDF.
又
∵∠DAE + ∠ABD = ∠DBF + ∠ABD = 90°，
∴∠DAE = ∠DBF，
∴△ADE∽△BDF，
∴$\frac{BF}{AE} = \frac{BD}{AD} = \frac{\frac{4\sqrt{5}}{5}}{\frac{8\sqrt{5}}{5}} = \frac{1}{2}$，则$BF = \frac{1}{2}AE = \frac{1}{2}x$，
∴$y = S_{四边形BEDF} = S_{\triangle BDE} + S_{\triangle BDF} = \frac{1}{2}(4 - x) × \frac{8}{5} + \frac{1}{2} × \frac{1}{2}x × \frac{4}{5} = -\frac{3}{5}x + \frac{16}{5}(0 ＜ x ＜ 4)$.
2. 如图，在$\triangle ABC$中，$\angle ACB = 90^{\circ}$，$CD\perp AB$，垂足为$D$，$E$是$BC$边上一点，连接$AE$交$CD$于点$F$，且$AC^{2}=CD· AE$.
(1)求证：$\angle CEA = \angle CAB$；
(2)作$CG\perp AE$，垂足为$G$，延长$CG$交$AB$于点$M$，求证：$AE· CM = AB· CF$.
(1)求证：$\angle CEA = \angle CAB$；
(2)作$CG\perp AE$，垂足为$G$，延长$CG$交$AB$于点$M$，求证：$AE· CM = AB· CF$.
答案：
2.证明:
(1)
∵$AC^{2} = CD· AE$，
∴$\frac{AE}{AC} = \frac{AC}{CD}$.
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)可知∠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，
∴$\frac{AB}{AE} = \frac{CD}{AD}$，$AF = CF$.
∵$CD\bot AB$，$CG\bot AE$，
∴∠CDM = ∠ADF = ∠CGF = 90°.
又
∵∠AFD = ∠CFG，
∴∠DCM = ∠DAF，
∴△DCM∽△DAF，
∴$\frac{CM}{AF} = \frac{CD}{AD}$，
∴$\frac{CM}{AF} = \frac{AB}{AE}$.
又
∵$AF = CF$，
∴$\frac{CM}{CF} = \frac{AB}{AE}$，
∴$AE· CM = AB· CF$.
2.证明:
(1)
∵$AC^{2} = CD· AE$，
∴$\frac{AE}{AC} = \frac{AC}{CD}$.
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)可知∠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，
∴$\frac{AB}{AE} = \frac{CD}{AD}$，$AF = CF$.
∵$CD\bot AB$，$CG\bot AE$，
∴∠CDM = ∠ADF = ∠CGF = 90°.
又
∵∠AFD = ∠CFG，
∴∠DCM = ∠DAF，
∴△DCM∽△DAF，
∴$\frac{CM}{AF} = \frac{CD}{AD}$，
∴$\frac{CM}{AF} = \frac{AB}{AE}$.
又
∵$AF = CF$，
∴$\frac{CM}{CF} = \frac{AB}{AE}$，
∴$AE· CM = AB· CF$.
3. 如图，在$Rt\triangle ACB$中，$\angle ACB = 90^{\circ}$，$AC = BC = 6$，$CD = 2$，$CH\perp BD$于点$H$，$O$是$AB$的中点，连接$OH$，求$OH$的长度.
答案：
3.解:在BH上截取$BE = CH$，连接CO，OE，如答图所示.
∵∠ACB = 90°，
∴在Rt△BCD中，$BC = 6$，$CD = 2$，
由勾股定理，得$BD = \sqrt{BC^{2} + CD^{2}} = \sqrt{6^{2} + 2^{2}} = 2\sqrt{10}$.
∵∠ACB = 90°，$CH\bot BD$，
∴∠DCH + ∠BCH = 90°，∠DBC + ∠BCH = 90°，
∴∠DCH = ∠DBC，又
∵∠CDH = ∠BDC，
∴△CDH∽△BDC，
∴$\frac{CH}{BC} = \frac{DH}{CD} = \frac{CD}{BD}$，
∴$\frac{CH}{6} = \frac{DH}{2} = \frac{2}{2\sqrt{10}}$，
∴$CH = \frac{3\sqrt{10}}{5}$，$DH = \frac{\sqrt{10}}{5}$，
∴$BE = CH = \frac{3\sqrt{10}}{5}$，
∴$EH = BD - BE - DH = 2\sqrt{10} - \frac{3\sqrt{10}}{5} - \frac{\sqrt{10}}{5} = \frac{6\sqrt{10}}{5}$.
在Rt△ACB中，∠ACB = 90°，$AC = BC = 6$，O是AB的中点，
∴$OC\bot AB$，$OB = OA = OC$，∠ABC = 45°，∠ACO = 45°，
∴∠EBO + ∠DBC = 45°，∠DCH + ∠HCO = 45°.
∵∠DCH = ∠DBC，
∴∠EBO = ∠HCO.
在△EBO和△HCO中，$\begin{cases}OB = OC,\\∠EBO = ∠HCO,\\BE = CH,\end{cases}$
∴△EBO≌△HCO(SAS)，
∴$OE = OH$，∠EOB = ∠HOC.
∵$OC\bot AB$，
∴∠COE + ∠EOB = 90°，
∴∠COE + ∠HOC = 90°，即∠EOH = 90°，
∴△EOH是等腰直角三角形，由勾股定理，得$EH = \sqrt{OH^{2} + OE^{2}} = \sqrt{2}OH$，
∴$OH = \frac{\sqrt{2}}{2}EH = \frac{\sqrt{2}}{2} × \frac{6\sqrt{10}}{5} = \frac{6\sqrt{5}}{5}$.
3.解:在BH上截取$BE = CH$，连接CO，OE，如答图所示.
∵∠ACB = 90°，
∴在Rt△BCD中，$BC = 6$，$CD = 2$，
由勾股定理，得$BD = \sqrt{BC^{2} + CD^{2}} = \sqrt{6^{2} + 2^{2}} = 2\sqrt{10}$.
∵∠ACB = 90°，$CH\bot BD$，
∴∠DCH + ∠BCH = 90°，∠DBC + ∠BCH = 90°，
∴∠DCH = ∠DBC，又
∵∠CDH = ∠BDC，
∴△CDH∽△BDC，
∴$\frac{CH}{BC} = \frac{DH}{CD} = \frac{CD}{BD}$，
∴$\frac{CH}{6} = \frac{DH}{2} = \frac{2}{2\sqrt{10}}$，
∴$CH = \frac{3\sqrt{10}}{5}$，$DH = \frac{\sqrt{10}}{5}$，
∴$BE = CH = \frac{3\sqrt{10}}{5}$，
∴$EH = BD - BE - DH = 2\sqrt{10} - \frac{3\sqrt{10}}{5} - \frac{\sqrt{10}}{5} = \frac{6\sqrt{10}}{5}$.
在Rt△ACB中，∠ACB = 90°，$AC = BC = 6$，O是AB的中点，
∴$OC\bot AB$，$OB = OA = OC$，∠ABC = 45°，∠ACO = 45°，
∴∠EBO + ∠DBC = 45°，∠DCH + ∠HCO = 45°.
∵∠DCH = ∠DBC，
∴∠EBO = ∠HCO.
在△EBO和△HCO中，$\begin{cases}OB = OC,\\∠EBO = ∠HCO,\\BE = CH,\end{cases}$
∴△EBO≌△HCO(SAS)，
∴$OE = OH$，∠EOB = ∠HOC.
∵$OC\bot AB$，
∴∠COE + ∠EOB = 90°，
∴∠COE + ∠HOC = 90°，即∠EOH = 90°，
∴△EOH是等腰直角三角形，由勾股定理，得$EH = \sqrt{OH^{2} + OE^{2}} = \sqrt{2}OH$，
∴$OH = \frac{\sqrt{2}}{2}EH = \frac{\sqrt{2}}{2} × \frac{6\sqrt{10}}{5} = \frac{6\sqrt{5}}{5}$.
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