$证明:连接AB\ $ $在Rt△ABC和Rt△BAD中$ $\begin{cases}{ AB=BA }\ \\ { BC=AD } \end{cases}$ $∴Rt△ABC≌Rt△BAD(HL),∴AC=BD$
$证明:连接AD$ $∵DE⊥AB,DF⊥AC,∴∠DEA=∠DFA=90°$ $在Rt△ADE和Rt△ADF中$ $\begin{cases}{ AD=AD }\ \\ { DE=DF } \end{cases}$ $∴Rt△ADE≌Rt△ADF(HL),∴AE=AF $
$证明:∵AB=AC,AE=AF,∴BE=CF$ $在△BDE和△CDF中$ $\begin{cases}{ DE=DF }\ \\ { ∠DEB=∠DFC } \\{ BE=CF} \end{cases}$ $∴△BDE≌△CDF(SAS)$ $\ ∴DB=DC$
$证明:∵DE⊥AB于点E,DF⊥AC于点F\ $ $∴∠E=∠DFC=∠DFA=90°$ $在Rt△EBD和Rt△FCD中$ $\begin{cases}{ BD=CD }\ \\ { BE=CF } \end{cases}$ $∴Rt△EBD≌Rt△FCD(HL),∴DE=DF$ $在Rt△AED和Rt△AFD中$ $\begin{cases}{ AD=AD }\ \\ { DE=DF } \end{cases}$ $∴Rt△AED≌Rt△AFD(HL) $
$解:∵Rt△AED≌Rt△AFD,∴AE=AF\ $ $∴AF=AB+BE=12+BE\ $ $∵AC=AF+FC,∴AC=AB+BE+FC\ $ $∴18=12+BE+CF\ $ $∵BE=CF,∴18=12+2BE,∴BE=3 $
$证明:∵DE⊥AB,∠ACB=90°\ $ $∴∠AED=∠AEF=∠ACB=90°$ $在Rt△ACF与Rt△AEF中$ $\begin{cases}{ AF=AF }\ \\ { AC=AE } \end{cases}$ $∴Rt△ACF≌Rt△AEF(HL),∴CF=EF$ $在Rt△ADE与Rt△ABC中\ $ $\begin{cases}{ AD=AB }\ \\ { AE=AC } \end{cases}$ $∴Rt△ADE≌Rt△ABC(HL),∴DE=BC\ $ $∵DF=DE+EF,∴DF=BC+CF $
$解:BC=CF+DF,证明如下:\ $ $连接AF$ $在Rt△ABC与Rt△ADE中$ $\begin{cases}{ AB=AD }\ \\ {AC=AE\ } \end{cases}$ $∴Rt△ABC≌Rt△ADE(HL),∴BC=DE\ $ $∵∠ACB=90°,∴∠ACF=90°=∠AED$ $在Rt△ACF与Rt△AEF中$ $\begin{cases}{ AF=AF }\ \\ { AC=AE } \end{cases}$ $∴Rt△ACF≌△AEF(HL)$ $∴CF=EF$ $∵DE=EF+DF,∴BC=CF+DF $
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