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1. 如图，在$△ ABC$中，点$D$在边$AB$上，点$E$在线段$CD$上，且$∠ ACD=∠ B=∠ BAE$.
(1)求证：$\frac{AD}{BC}=\frac{DE}{AC}$；
(2)当点$E$为$CD$中点时，求证：$\frac{AE^2}{CE^2}=\frac{AB}{AD}$.
(1)求证：$\frac{AD}{BC}=\frac{DE}{AC}$；
(2)当点$E$为$CD$中点时，求证：$\frac{AE^2}{CE^2}=\frac{AB}{AD}$.
答案
(1) $\because ∠ACD=∠B,∠BAC=∠CAD, \therefore △ADC ∽ △ACB. \because ∠ACD=∠BAE,∠ADE=∠CDA, \therefore △ADE ∽ △CDA. \therefore △ADE ∽ △BCA. \therefore \frac{AD}{BC}=\frac{DE}{AC}.$
(2) $\because △ADE∽△BCA, \therefore \frac{AE}{AB}=\frac{DE}{AC}, 即 \frac{AE}{DE} = \frac{AB}{AC}. \because △ADE ∽ △CDA, \therefore \frac{AE}{AC}=\frac{DE}{AD}, 即 \frac{AE}{DE} = \frac{AC}{AD}. \therefore \frac{AE^2}{DE^2}=\frac{AB}{AC} · \frac{AC}{AD}=\frac{AB}{AD}. \because 点 E 为 CD 中点, \therefore DE=CE. \therefore \frac{AE^2}{CE^2}=\frac{AB}{AD}.$
(2) $\because △ADE∽△BCA, \therefore \frac{AE}{AB}=\frac{DE}{AC}, 即 \frac{AE}{DE} = \frac{AB}{AC}. \because △ADE ∽ △CDA, \therefore \frac{AE}{AC}=\frac{DE}{AD}, 即 \frac{AE}{DE} = \frac{AC}{AD}. \therefore \frac{AE^2}{DE^2}=\frac{AB}{AC} · \frac{AC}{AD}=\frac{AB}{AD}. \because 点 E 为 CD 中点, \therefore DE=CE. \therefore \frac{AE^2}{CE^2}=\frac{AB}{AD}.$
2. 如图,在梯形$ABCD$中,$AB// CD$,$∠ D=90°$,$AD=CD=2$,点$E$在边$AD$上(不与点$A$、$D$重合),$∠ CEB=45°$,$EB$与对角线$AC$相交于点$F$,设$DE=x$.
(1)用含$x$的代数式表示线段$CF$的长;
(2)如果把$△ CAE$的周长记作$C_{△ CAE}$,$△ BAF$的周长记作$C_{△ BAF}$,设$\frac{C_{△ CAE}}{C_{△ BAF}}=y$,求$y$关于$x$的函数关系式,并写出它的自变量的取值范围;
(3)当$\frac{AE}{AB}=\frac{3}{5}$时,求$AB$的长.
(1)用含$x$的代数式表示线段$CF$的长;
(2)如果把$△ CAE$的周长记作$C_{△ CAE}$,$△ BAF$的周长记作$C_{△ BAF}$,设$\frac{C_{△ CAE}}{C_{△ BAF}}=y$,求$y$关于$x$的函数关系式,并写出它的自变量的取值范围;
(3)当$\frac{AE}{AB}=\frac{3}{5}$时,求$AB$的长.
答案
(1) $\because ∠D=90°,AD=CD, \therefore ∠DAF=45°. \because ∠CEB=45°, \therefore ∠DAF = ∠CEB. \because ∠ECA = ∠FCE, \therefore △ECA∽△FCE. \therefore \frac{EC}{FC}=\frac{CA}{CE}. \because ∠D=90°,DE=x. \therefore AC=\sqrt{AD^2+DC^2}=2\sqrt{2}. \because DE=x, EC=\sqrt{DE^2+DC^2}=\sqrt{x^2+4}, \therefore \frac{\sqrt{x^2+4}}{FC}=\frac{2\sqrt{2}}{\sqrt{x^2+4}}. \therefore FC=\frac{x^2+4}{2\sqrt{2}}=\frac{4\sqrt{2}+\sqrt{2}x^2}{4}.$
(2) $\because AB // CD, ∠D = 90°, \therefore ∠EAC=∠FAB=45°. \because ∠CEF = ∠FAB, ∠EFC = ∠AFB, \therefore ∠ECA = ∠FBA. \therefore △CAE ∽ △BAF. \therefore \frac{C_{△CAE}}{C_{△BAF}}=\frac{AE}{AF}. \because AC=2\sqrt{2}, FC = \frac{4\sqrt{2}+\sqrt{2}x^2}{4}, \therefore AF = \frac{4\sqrt{2}-\sqrt{2}x^2}{4}. \because DE=x,AD=2, \therefore CD = 2 - x. \therefore y=\frac{C_{△CAE}}{C_{△BAF}} = \frac{AE}{AF} = \frac{2-x}{\frac{4\sqrt{2}-\sqrt{2}x^2}{4}}=\frac{2\sqrt{2}}{2+x}(0＜x＜2).$
(3) $\because ∠CEF=∠FAB,∠EFC =∠AFB, \therefore △EFC∽△AFB. \therefore \frac{EF}{AF}=\frac{FC}{BF}. \because ∠EFA = ∠CFB, \therefore △EFA ∽△CFB. \therefore ∠EAF=∠CBF=45°. \therefore CE = CB.$
过点 C 作 $CH ⊥ AB$,垂足为点 H,则 $AD=CD=CH=AH=2$.
$\therefore Rt△CDE≌Rt△CHB. \therefore DE =BH=x.$
$Rt△EAB$ 中,$\frac{AE}{AB}=\frac{2-x}{2+x}=\frac{3}{5}.$
解得 $x=\frac{1}{2}. \therefore AB=AH+BH=\frac{5}{2}.$
(2) $\because AB // CD, ∠D = 90°, \therefore ∠EAC=∠FAB=45°. \because ∠CEF = ∠FAB, ∠EFC = ∠AFB, \therefore ∠ECA = ∠FBA. \therefore △CAE ∽ △BAF. \therefore \frac{C_{△CAE}}{C_{△BAF}}=\frac{AE}{AF}. \because AC=2\sqrt{2}, FC = \frac{4\sqrt{2}+\sqrt{2}x^2}{4}, \therefore AF = \frac{4\sqrt{2}-\sqrt{2}x^2}{4}. \because DE=x,AD=2, \therefore CD = 2 - x. \therefore y=\frac{C_{△CAE}}{C_{△BAF}} = \frac{AE}{AF} = \frac{2-x}{\frac{4\sqrt{2}-\sqrt{2}x^2}{4}}=\frac{2\sqrt{2}}{2+x}(0＜x＜2).$
(3) $\because ∠CEF=∠FAB,∠EFC =∠AFB, \therefore △EFC∽△AFB. \therefore \frac{EF}{AF}=\frac{FC}{BF}. \because ∠EFA = ∠CFB, \therefore △EFA ∽△CFB. \therefore ∠EAF=∠CBF=45°. \therefore CE = CB.$
过点 C 作 $CH ⊥ AB$,垂足为点 H,则 $AD=CD=CH=AH=2$.
$\therefore Rt△CDE≌Rt△CHB. \therefore DE =BH=x.$
$Rt△EAB$ 中,$\frac{AE}{AB}=\frac{2-x}{2+x}=\frac{3}{5}.$
解得 $x=\frac{1}{2}. \therefore AB=AH+BH=\frac{5}{2}.$
