【例1】如图,$AC⊥BC$于点C,$AD⊥BD$于点D,$AD= BC,CE⊥AB,DF⊥AB$,垂足分别是E,F. 求证$CE= DF.$

证明:$\because AC⊥BC,AD⊥BD,$
$\therefore ∠ACB= ∠BDA= 90^{\circ }.$
在$Rt△ABC和Rt△BAD$中,
$\left\{\begin{array}{l} AB= BA,\\ BC= AD,\end{array} \right.$
$\therefore Rt△ABC\cong Rt△BAD$().
$\therefore ∠CBE= ∠DAF.$
$\because CE⊥AB,DF⊥AB,$
$\therefore ∠CEB= ∠DFA= 90^{\circ }.$
在$△BCE和△ADF$中,
$\left\{\begin{array}{l} ∠CEB= ∠DFA,\\ ∠CBE= ∠DAF,\\ BC= AD,\end{array}\right.$
$\therefore △BCE\cong △ADF$().
$\therefore CE= DF.$

【分析】先用“HL”判定$Rt△ABC\cong Rt△BAD$,得$∠CBE= ∠DAF$,再用“AAS”判定$△BCE\cong △ADF$,从而推出$CE= DF.$
证明:$\because AC⊥BC,AD⊥BD,$
$\therefore ∠ACB= ∠BDA= 90^{\circ }.$
在$Rt△ABC和Rt△BAD$中,
$\left\{\begin{array}{l} AB= BA,\\ BC= AD,\end{array} \right. $
$\therefore Rt△ABC\cong Rt△BAD(HL).$
$\therefore ∠CBE= ∠DAF.$
$\because CE⊥AB,DF⊥AB,$
$\therefore ∠CEB= ∠DFA= 90^{\circ }.$
在$△BCE和△ADF$中,
$\left\{\begin{array}{l} ∠CEB= ∠DFA,\\ ∠CBE= ∠DAF,\\ BC= AD,\end{array} \right. $
$\therefore △BCE\cong △ADF(AAS).$
$\therefore CE= DF.$

【例2】如图,AD,AF分别是钝角三角形ABC和钝角三角形ABE的高,$AD= AF,AC= AE$. 求证$BC= BE.$

证明:$\because AD,AF$分别是钝角三角形ABC和钝角三角形ABE的高,
$\therefore ∠ADB= ∠AFB= 90^{\circ }.$
在$Rt△ABD$和$Rt△ABF$中,
$\left\{\begin{array}{l} AB= AB,\\ AD= AF,\end{array}\right.$
$\therefore Rt△ABD\cong Rt△ABF$().
$\therefore DB= FB.$
在$Rt△ADC$和$Rt△AFE$中,
$\left\{\begin{array}{l} AC= AE,\\ AD= AF,\end{array}\right.$
$\therefore Rt△ADC\cong Rt△AFE$().
$\therefore DC= FE.$
$\therefore DB-DC= FB-FE$,即$BC= BE.$

【分析】用“HL”先判定$Rt△ABD\cong Rt△ABF$,再判定$Rt△ADC\cong Rt△AFE$,然后得出$BC= BE.$
证明:$\because AD,AF$分别是钝角三角形ABC和钝角三角形ABE的高,
$\therefore ∠ADB= ∠AFB= 90^{\circ }.$
在$Rt△ABD和Rt△ABF$中,
$\left\{\begin{array}{l} AB= AB,\\ AD= AF,\end{array} \right. $
$\therefore Rt△ABD\cong Rt△ABF(HL).$
$\therefore DB= FB.$
在$Rt△ADC和Rt△AFE$中,
$\left\{\begin{array}{l} AC= AE,\\ AD= AF,\end{array} \right. $
$\therefore Rt△ADC\cong Rt△AFE(HL).$
$\therefore DC= FE.$
$\therefore DB-DC= FB-FE$,即$BC= BE.$

一、选择题
1. 如图,$∠A= ∠D= 90^{\circ },AC= DB$,则$△ABC\cong △DCB$的理由是()


A.HL
B.ASA
C.AAS
D.SAS

A