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1. 如图，$AB = 8$，$BD \perp AB$，$AC \perp AB$，且$AC = 3$.$E$是线段$AB$上一动点，过点$E$作$CE$的垂线，交射线$BD$于点$F$，则$BF$的长的最大值是
$\frac{16}{3}$
.
答案：
1. $\frac{16}{3}$ 点拨：设$AE=x$，$\because AB=8$，$\therefore BE=8 - x$。
$\because BD\perp AB$，$AC\perp AB$，$CE\perp EF$，
$\therefore \angle CAE=\angle CEF=\angle EBF=90^{\circ}$，
$\therefore \angle ACE+\angle AEC=90^{\circ}$，$\angle BEF+\angle AEC=90^{\circ}$，
$\therefore \angle ACE=\angle BEF$，$\therefore \triangle ACE\sim \triangle BEF$，
$\therefore \frac{AE}{AC}=\frac{BF}{BE}$，即$\frac{x}{3}=\frac{BF}{8 - x}$，
$\therefore BF=-\frac{1}{3}x^{2}+\frac{8}{3}x=-\frac{1}{3}(x - 4)^{2}+\frac{16}{3}$，
$\therefore$当$x = 4$时，$BF$的长取得最大值，最大值为$\frac{16}{3}$。
$\because BD\perp AB$，$AC\perp AB$，$CE\perp EF$，
$\therefore \angle CAE=\angle CEF=\angle EBF=90^{\circ}$，
$\therefore \angle ACE+\angle AEC=90^{\circ}$，$\angle BEF+\angle AEC=90^{\circ}$，
$\therefore \angle ACE=\angle BEF$，$\therefore \triangle ACE\sim \triangle BEF$，
$\therefore \frac{AE}{AC}=\frac{BF}{BE}$，即$\frac{x}{3}=\frac{BF}{8 - x}$，
$\therefore BF=-\frac{1}{3}x^{2}+\frac{8}{3}x=-\frac{1}{3}(x - 4)^{2}+\frac{16}{3}$，
$\therefore$当$x = 4$时，$BF$的长取得最大值，最大值为$\frac{16}{3}$。
2. 如图，在长方形$ABCD$中，内接三个大小相同的正方形，点$E,F,G,H$分别在边$AB$，$BC,CD,AD$上，若$AB = 12\ cm$，$BC = 10\ cm$，则每个小正方形的边长为
$2\sqrt{5}$
$cm$.
答案：
2. $2\sqrt{5}$ 点拨：$\because$四边形$ABCD$是矩形，$\therefore \angle A=\angle B=\angle C=90^{\circ}$。$\because \angle HEF=\angle EFG=90^{\circ}$，$\therefore \angle AEH=\angle BFE=90^{\circ}-\angle BEF$，$\angle BFE=\angle CGF=90^{\circ}-\angle CFG$；
在$\triangle AEH$和$\triangle BFE$中，$\begin{cases} \angle A=\angle B,\\ \angle AEH=\angle BFE,\\ HE=EF,\\ \end{cases}$
$\therefore \triangle AEH\cong \triangle BFE(AAS)$，$\therefore AE=BF$。$\because \angle B=\angle C$，$\angle BFE=\angle CGF$，$EF = 2FG$，
$\therefore \triangle BFE\sim \triangle CGF$，$\therefore \frac{BE}{CF}=\frac{EF}{FG}=2$，
$\therefore BE = 2CF$。$\because AB = 12cm$，$BC = 10cm$，$\therefore AE=BF=10 - CF$，$\therefore 10 - CF+2CF = 12$，
解得$CF = 2$，$\therefore BE = 4cm$，$BF = 8cm$，$\therefore EF=\sqrt{BE^{2}+BF^{2}}=\sqrt{4^{2}+8^{2}}=4\sqrt{5}(cm)$，$\therefore FG=\frac{1}{2}EF=2\sqrt{5}cm$，$\therefore$每个小正方形的边长为$2\sqrt{5}cm$。
在$\triangle AEH$和$\triangle BFE$中，$\begin{cases} \angle A=\angle B,\\ \angle AEH=\angle BFE,\\ HE=EF,\\ \end{cases}$
$\therefore \triangle AEH\cong \triangle BFE(AAS)$，$\therefore AE=BF$。$\because \angle B=\angle C$，$\angle BFE=\angle CGF$，$EF = 2FG$，
$\therefore \triangle BFE\sim \triangle CGF$，$\therefore \frac{BE}{CF}=\frac{EF}{FG}=2$，
$\therefore BE = 2CF$。$\because AB = 12cm$，$BC = 10cm$，$\therefore AE=BF=10 - CF$，$\therefore 10 - CF+2CF = 12$，
解得$CF = 2$，$\therefore BE = 4cm$，$BF = 8cm$，$\therefore EF=\sqrt{BE^{2}+BF^{2}}=\sqrt{4^{2}+8^{2}}=4\sqrt{5}(cm)$，$\therefore FG=\frac{1}{2}EF=2\sqrt{5}cm$，$\therefore$每个小正方形的边长为$2\sqrt{5}cm$。
3. 如图，在$Rt \bigtriangleup ABC$中，$\angle ACB = 90^{\circ}$，过点$B$作$BD \perp AB$，交$\angle ACB$的平分线于点$D$，$AB$与$CD$相交于点$E$.若$BE = \sqrt{10}$，$BD = 2\sqrt{10}$，求$AC$的长.
答案：
3. 解：如答图，作$EF\perp CB$于点$F$，$DL\perp CB$交$CB$的延长线于点$L$，则$\angle CFE=\angle EFB=\angle L=90^{\circ}$。$\because \angle ACB=90^{\circ}$，$CD$平分$\angle ACB$，$\therefore \angle BCD=\frac{1}{2}\angle ACB=45^{\circ}$，$\therefore \angle FEC=\angle LDC=\angle LCD=45^{\circ}$，$\therefore EF=CF$，$DL=CL$。$\because BD\perp AB$，$BE=\sqrt{10}$，$BD=2\sqrt{10}$，
$\therefore \angle ABD=90^{\circ}$，$\therefore \angle FBE=\angle LDB=90^{\circ}-\angle LBD$，
$\therefore \triangle FBE\sim \triangle LDB$，
$\therefore \frac{EF}{BL}=\frac{BF}{DL}=\frac{BE}{BD}=\frac{\sqrt{10}}{2\sqrt{10}}=\frac{1}{2}$，$\therefore BL = 2EF$，$DL = 2BF$，设$EF=CF=m$，则$BL = 2m$。
$\because DL=CL=2BF=BF+m+2m$，$\therefore BF = 3m$，
$\therefore BC=m + 3m=4m$。
$\because BE=\sqrt{EF^{2}+BF^{2}}=\sqrt{m^{2}+(3m)^{2}}=\sqrt{10m^{2}}=\sqrt{10}$，$\therefore EF=m = 1$。
$\because \angle ACB=\angle EFB=90^{\circ}$，$\angle ABC=\angle EBF$，
$\therefore \triangle ABC\sim \triangle EBF$，
$\therefore \frac{AC}{EF}=\frac{BC}{BF}=\frac{4m}{3m}=\frac{4}{3}$，$\therefore AC=\frac{4}{3}EF=\frac{4}{3}×1=\frac{4}{3}$。
3. 解：如答图，作$EF\perp CB$于点$F$，$DL\perp CB$交$CB$的延长线于点$L$，则$\angle CFE=\angle EFB=\angle L=90^{\circ}$。$\because \angle ACB=90^{\circ}$，$CD$平分$\angle ACB$，$\therefore \angle BCD=\frac{1}{2}\angle ACB=45^{\circ}$，$\therefore \angle FEC=\angle LDC=\angle LCD=45^{\circ}$，$\therefore EF=CF$，$DL=CL$。$\because BD\perp AB$，$BE=\sqrt{10}$，$BD=2\sqrt{10}$，
$\therefore \angle ABD=90^{\circ}$，$\therefore \angle FBE=\angle LDB=90^{\circ}-\angle LBD$，
$\therefore \triangle FBE\sim \triangle LDB$，
$\therefore \frac{EF}{BL}=\frac{BF}{DL}=\frac{BE}{BD}=\frac{\sqrt{10}}{2\sqrt{10}}=\frac{1}{2}$，$\therefore BL = 2EF$，$DL = 2BF$，设$EF=CF=m$，则$BL = 2m$。
$\because DL=CL=2BF=BF+m+2m$，$\therefore BF = 3m$，
$\therefore BC=m + 3m=4m$。
$\because BE=\sqrt{EF^{2}+BF^{2}}=\sqrt{m^{2}+(3m)^{2}}=\sqrt{10m^{2}}=\sqrt{10}$，$\therefore EF=m = 1$。
$\because \angle ACB=\angle EFB=90^{\circ}$，$\angle ABC=\angle EBF$，
$\therefore \triangle ABC\sim \triangle EBF$，
$\therefore \frac{AC}{EF}=\frac{BC}{BF}=\frac{4m}{3m}=\frac{4}{3}$，$\therefore AC=\frac{4}{3}EF=\frac{4}{3}×1=\frac{4}{3}$。
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