22. (8分)如图，在$□ ABCD$中，$E$是$AD$的中点，连接$CE$并延长，交$BA$的延长线于点$F$.
(1) 求证：$AF=AB$；
(2) 若$G$是线段$AF$上一点，满足$\angle FCG=\angle FCD$，$CG$交$AD$于点$H$，$AG=2$，$FG=6$，求$CH$的长.
答案：22.(1)$\because$四边形$ABCD$是平行四边形，$\therefore AD// BC$，$CD// AB$。$\therefore \angle D = \angle FAD$，$\angle DCE = \angle F$。$\because E$是$AD$的中点，$\therefore DE = AE$。$\therefore \triangle CDE\cong\triangle FAE$。$\therefore CE = FE$。$\because AE// BC$，$\therefore \frac{FA}{AB}=\frac{FE}{CE}=1$。$\therefore AF = AB$。(2)$\because AG = 2$，$FG = 6$，$\therefore AF = FG + AG = 6 + 2 = 8$。$\therefore AB = AF = 8$。$\because$四边形$ABCD$是平行四边形，$\therefore CD = AB = 8$，$DC// AB$。$\therefore \angle DCF = \angle F$。$\because \angle FCG = \angle FCD$，$\therefore \angle FCG = \angle F$。$\therefore CG = FG = 6$。$\because CD// AF$，$\therefore \triangle DCH\sim\triangle AGH$。$\therefore \frac{CD}{GA}=\frac{CH}{GH}$，即$\frac{8}{2}=\frac{6 - GH}{GH}$。$\therefore GH = 1.2$。$\therefore CH = CG - GH = 6 - 1.2 = 4.8$。
解析：
(1)证明：
∵四边形$ABCD$是平行四边形，
∴$AD// BC$，$CD// AB$，
∴$\angle D=\angle FAD$，$\angle DCE=\angle F$，
∵$E$是$AD$的中点，
∴$DE=AE$，
在$\triangle CDE$和$\triangle FAE$中，
$\left\{\begin{array}{l}\angle DCE=\angle F\\\angle D=\angle FAD\\DE=AE\end{array}\right.$，
∴$\triangle CDE\cong\triangle FAE(AAS)$，
∴$CE=FE$，
∵$AE// BC$，
∴$\frac{FA}{AB}=\frac{FE}{CE}=1$，
∴$AF=AB$。
(2)解：
∵$AG=2$，$FG=6$，
∴$AF=FG+AG=6+2=8$，
由(1)知$AF=AB$，
∴$AB=8$，
∵四边形$ABCD$是平行四边形，
∴$CD=AB=8$，$DC// AB$，
∴$\angle DCF=\angle F$，
∵$\angle FCG=\angle FCD$，
∴$\angle FCG=\angle F$，
∴$CG=FG=6$，
∵$CD// AF$，
∴$\triangle DCH\sim\triangle AGH$，
∴$\frac{CD}{GA}=\frac{CH}{GH}$，
设$GH=x$，则$CH=6-x$，
∴$\frac{8}{2}=\frac{6-x}{x}$，
解得$x=1.2$，
∴$CH=6-1.2=4.8$。
23. (8分)如图，$\triangle ABC$是$\odot O$的内接三角形，$AB$是$\odot O$的直径，$AC=\sqrt{5}$，$BC=2\sqrt{5}$，点$F$在$AB$上，连接$CF$并延长，交$\odot O$于点$D$，连接$BD$，过点$B$作$BE\perp CD$，垂足为$E$.
(1) 求证：$\triangle DBE \backsim \triangle ABC$；
(2) 若$AF=2$，求$DE$的长.
答案：23.(1)$\because AB$是$\odot O$的直径，$\therefore \angle ACB = 90^{\circ}$。$\because BE\perp CD$，$\therefore \angle BED = 90^{\circ}=\angle BCA$。又$\because \overset{\frown}{BC}=\overset{\frown}{BC}$，$\therefore \angle D = \angle A$。$\therefore \triangle DBE\sim\triangle ABC$。(2)过点$C$作$CG\perp AB$，垂足为$G$。$\because \angle ACB = 90^{\circ}$，$AC = \sqrt{5}$，$BC = 2\sqrt{5}$，$\therefore AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{(\sqrt{5})^{2}+(2\sqrt{5})^{2}}=5$。$\because CG\perp AB$，$\therefore \angle AGC = 90^{\circ}=\angle ACB$。又$\because \angle A = \angle A$，$\therefore \triangle ACG\sim\triangle ABC$。$\therefore \frac{AG}{AC}=\frac{AC}{AB}$。$\therefore \frac{AG}{\sqrt{5}}=\frac{\sqrt{5}}{5}$。$\therefore AG = 1$。$\because AF = 2$，$\therefore FG = AG = 1$。$\therefore AC = FC$。$\therefore \angle A = \angle CFA = \angle BFD = \angle D$。$\therefore BD = BF = AB - AF = 5 - 2 = 3$。$\because \triangle DBE\sim\triangle ABC$，$\therefore \frac{BD}{BA}=\frac{DE}{AC}$。$\therefore \frac{3}{5}=\frac{DE}{\sqrt{5}}$。$\therefore DE=\frac{3\sqrt{5}}{5}$。
解析：
(1) 证明：
∵ $AB$ 是 $\odot O$ 的直径，
∴ $\angle ACB = 90°$。
∵ $BE \perp CD$，
∴ $\angle BED = 90° = \angle ACB$。
∵ $\overset{\frown}{BC} = \overset{\frown}{BC}$，
∴ $\angle D = \angle A$。
∴ $\triangle DBE \backsim \triangle ABC$。
(2) 解：
∵ $\angle ACB = 90°$，$AC = \sqrt{5}$，$BC = 2\sqrt{5}$，
∴ $AB = \sqrt{AC^2 + BC^2} = \sqrt{(\sqrt{5})^2 + (2\sqrt{5})^2} = 5$。
过点 $C$ 作 $CG \perp AB$ 于 $G$，
∵ $\angle AGC = 90° = \angle ACB$，$\angle A = \angle A$，
∴ $\triangle ACG \backsim \triangle ABC$。
∴ $\frac{AG}{AC} = \frac{AC}{AB}$，即 $\frac{AG}{\sqrt{5}} = \frac{\sqrt{5}}{5}$，
解得 $AG = 1$。
∵ $AF = 2$，
∴ $FG = AF - AG = 2 - 1 = 1$，
∴ $AG = FG$，即 $CG$ 垂直平分 $AF$，
∴ $AC = FC$，$\angle A = \angle CFA$。
∵ $\angle CFA = \angle BFD$，$\angle A = \angle D$，
∴ $\angle BFD = \angle D$，
∴ $BD = BF = AB - AF = 5 - 2 = 3$。
∵ $\triangle DBE \backsim \triangle ABC$，
∴ $\frac{BD}{BA} = \frac{DE}{AC}$，即 $\frac{3}{5} = \frac{DE}{\sqrt{5}}$，
解得 $DE = \frac{3\sqrt{5}}{5}$。
24. (10分)
(1) 如图①，$\triangle ABC \backsim \triangle ADE$，求证：$\triangle ABD \backsim \triangle ACE$.
(2) 如图②，在$\triangle ABC$和$\triangle ADE$中，$\angle BAC=\angle DAE=90°$，$\angle ABC=\angle ADE=30°$，$AC$与$DE$相交于点$F$，点$D$在边$BC$上，$\frac{AD}{BD}=\sqrt{3}$.求$\frac{DF}{CF}$的值.
(3) 如图③，$D$是$\triangle ABC$内一点，$\angle BAD=\angle CBD=30°$，$\angle BDC=90°$，$AD=\sqrt{5}$，$AC=2\sqrt{3}$.求$AB$的长.
答案：24.(1)$\because \triangle ABC\sim\triangle ADE$，$\therefore \frac{AB}{AD}=\frac{AC}{AE}$，$\angle BAC = \angle DAE$。$\therefore \frac{AB}{AC}=\frac{AD}{AE}$，$\angle BAC - \angle DAC = \angle DAE - \angle DAC$，即$\angle BAD = \angle CAE$。$\therefore \triangle ABD\sim\triangle ACE$。(2)连接$EC$。$\because \angle BAC = \angle DAE = 90^{\circ}$，$\angle ABC = \angle ADE = 30^{\circ}$，$\therefore \triangle ABC\sim\triangle ADE$。由(1)，同理，得$\triangle ABD\sim\triangle ACE$。$\therefore \frac{AD}{AE}=\frac{BD}{CE}$，$\angle ABD = \angle ACE = \angle ADE$。$\therefore \frac{AD}{BD}=\frac{AE}{CE}=\sqrt{3}$。在$Rt\triangle ADE$中，$\angle ADE = 30^{\circ}$，$\therefore$易得$\frac{AD}{AE}=\sqrt{3}$。$\therefore \frac{AD}{EC}=\frac{AD}{AE}· \frac{AE}{CE}=\sqrt{3}×\sqrt{3}=3$。$\because \angle ADF = \angle ECF = 30^{\circ}$，$\angle AFD = \angle EFC$，$\therefore \triangle ADF\sim\triangle ECF$。$\therefore \frac{DF}{CF}=\frac{AD}{EC}=3$。(3)如图，过点$A$作$AB$的垂线，过点$D$作$AD$的垂线，两垂线交于点$M$，连接$BM$，则$\angle BAM = \angle ADM = 90^{\circ}$。$\because \angle BAD = 30^{\circ}$，$\therefore \angle DAM = 60^{\circ}$。$\therefore \angle AMD = 30^{\circ}$。$\therefore$在$Rt\triangle ADM$中，$AM = 2AD = 2\sqrt{5}$，$DM = \sqrt{AM^{2}-AD^{2}}=\sqrt{(2\sqrt{5})^{2}-(\sqrt{5})^{2}}=\sqrt{15}$。$\because \angle ADM = \angle BDC = 90^{\circ}$，$\angle AMD = \angle CBD = 30^{\circ}$，$\therefore \triangle ADM\sim\triangle CDB$。$\therefore \frac{MD}{BD}=\frac{DA}{CD}$。$\therefore \frac{BD}{CD}=\frac{DM}{DA}$。又$\because \angle BDC = \angle MDA$，$\therefore \angle BDC + \angle CDM = \angle MDA + \angle CDM$，即$\angle BDM = \angle CDA$。$\therefore \triangle BDM\sim\triangle CDA$。$\therefore \frac{BM}{CA}=\frac{DM}{DA}=\frac{\sqrt{15}}{\sqrt{5}}=\sqrt{3}$。$\because AC = 2\sqrt{3}$，$\therefore BM = \sqrt{3}AC=\sqrt{3}×2\sqrt{3}=6$。$\therefore$在$Rt\triangle ABM$中，$AB=\sqrt{BM^{2}-AM^{2}}=\sqrt{6^{2}-(2\sqrt{5})^{2}}=4$。
