2. 如圖①,在$Rt\triangle ABC$中,$\angle ACB = 90^{\circ}$,$AC = BC$,直線$l經(jīng)過點(diǎn)C$,過點(diǎn)$A作AD\perp l$,垂足為$D$,過點(diǎn)$B作BE\perp l$,垂足為$E$.
(1)求證:$\triangle ADC\cong\triangle CEB$;
(2)若$AD = 5$,$DE = 13$,求$BE$的長(zhǎng);
(3)如圖②,延長(zhǎng)$AD至點(diǎn)F$,連接$CF$,過點(diǎn)$C作CG\perp CF$,且$CG = CF$,連接$BG交直線l于點(diǎn)H$,若$S_{\triangle CGH}= 30$,$CD = 10$,則$AF= $____.
答案：(1)證明:$\because AD⊥DE,BE⊥DE,$
$\therefore ∠ADC=∠CEB=∠ACB=90^{\circ }.$
$\therefore ∠DAC+∠DCA=∠ECB+∠DCA=90^{\circ },$
$\therefore ∠DAC=∠ECB.$
在$△ADC$和$△CEB$中,
$\left\{\begin{array}{l} ∠ADC=∠CEB,\\ ∠DAC=∠ECB,\\ AC=CB,\end{array}\right.$
$\therefore △ADC\cong △CEB(AAS).$
(2)解:$\because △ADC\cong △CEB,AD=5,$
$\therefore AD=CE=5,$$CD=BE.$
$\because DE=13,\therefore BE=CD=13 - 5=8.$
(3)12 點(diǎn)撥:過點(diǎn)G作$GM⊥l$于點(diǎn)M,如答圖.
$\because CG⊥CF,$
$\therefore ∠CFD+∠DCF=∠DCF+∠MCG=90^{\circ },$
$\therefore ∠CFD=∠MCG.$
$\because ∠CDF=∠CMG=90^{\circ },CF=CG,$
$\therefore △CDF\cong △GMC(AAS),$
$\therefore CD=GM=10,DF=MC.$
$\because S_{△CGH}=30,$
$\therefore \frac {1}{2}CH\cdot GM=30,\therefore CH=\frac {2×30}{GM}=\frac {60}{10}=6.$
$\because △ACD\cong △CBE,\therefore CD=BE=10,AD=CE,$
$\therefore BE=GM.$
$\because ∠BEH=∠GMH=90^{\circ },∠BHE=∠GHM,$
$\therefore △BEH\cong △GMH(AAS),\therefore EH=MH,$
$\therefore AF=AD+DF=CE+CM=CH - EH+CH+EH=2CH=2×6=12.$
