答案：21.解:(1)在$\mathrm{Rt}△MNB$中,
$\because BM=75\ \mathrm{m}$,$MN=60\ \mathrm{m}$,
$\therefore BN=\sqrt{BM^2-MN^2}=\sqrt{75^2-60^2}=45(\mathrm{m})$,
$\therefore AN=AB-BN=125-45=80(\mathrm{m})$,
$\therefore$在$\mathrm{Rt}△AMN$中,$AM=\sqrt{AN^2+MN^2}=\sqrt{80^2+60^2}=100(\mathrm{m})$.
$\therefore AM+BM=100+75=175(\mathrm{m})$,
$\therefore$供水点$M$到喷泉$A,B$需要铺设的管道总长为$175\ \mathrm{m}$.
(2)$\because AB=125\ \mathrm{m}$,$AM=100\ \mathrm{m}$,$BM=75\ \mathrm{m}$,
$\therefore AB^2=BM^2+AM^2$,
$\therefore △ABM$是直角三角形,
$\therefore BM⊥AC$,$\therefore$喷泉$B$到小路$AC$的最短距离是$75\ \mathrm{m}$.

答案：
22.解:如答图,连接$AC$.

(1)$\because AB⊥BC$,$\therefore ∠ABC=90°$.
$\because AB=9\ \mathrm{m}$,$BC=12\ \mathrm{m}$,
$\therefore AC=\sqrt{AB^2+BC^2}=\sqrt{9^2+12^2}=15(\mathrm{m})$.
$\because CD=17\ \mathrm{m}$,$AD=8\ \mathrm{m}$,
$\therefore AD^2+AC^2=8^2+15^2=289=17^2=CD^2$,
$\therefore △DAC$为直角三角形,$∠DAC=90°$,
$\therefore$需要绿化的空地$ABCD$的面积为$S_{△ABC}+S_{△DAC}=\dfrac{1}{2}AB·BC+\dfrac{1}{2}AD·AC=\dfrac{1}{2}×9×12+\dfrac{1}{2}×8×15=114(\mathrm{m}^2)$.
(2)$\because AE⊥CD$,$∠DAC=90°$,
$\therefore S_{△DAC}=\dfrac{1}{2}AD·AC=\dfrac{1}{2}CD·AE$,
$\therefore 8×15=17×AE$,解得$AE=\dfrac{120}{17}(\mathrm{m})$.

答案：
23.(1)证明:如答图①,延长$BD$交$AF$于点$M$.

$\because AF⊥EF$,$\therefore ∠AFE=90°$.
在$△BCD$和$△FCE$中,$\begin{cases} BC=FC, \\ ∠BCD=∠FCE, \\ DC=EC, \end{cases}$
$\therefore △BCD≌△FCE(\mathrm{SAS})$,$\therefore ∠CBD=∠CFE$,
$\therefore BD// EF$,$\therefore ∠AMB=∠AFE=90°$,$\therefore BD⊥AF$.
(2)解:补全图形如答图②,$CH=CD$.
理由如下:
如答图②,延长$BC$至点$F$,使$CF=BC$,连接$AF,EF$.
$\because ∠ACB=90°$,$\therefore AC⊥BF$,$\therefore AB=AF$.
由(1)可知$BH// EF$,$BD=EF$.
$\because AB^2=AE^2+BD^2$,$\therefore AF^2=AE^2+EF^2$,
$\therefore ∠AEF=90°$.
$\because BH// EF$,$\therefore ∠BHE=90°$.
又$\because CE=DC$,$\therefore CH=\dfrac{1}{2}DE=CD$.