【例1】如图,在四边形ABCD中,$AB// DC$,$AD// BC$.求证$\triangle ABC\cong \triangle CDA$.

证明:$\because AB// DC$,$AD// BC$,
$\therefore ∠1= ∠2$,$∠3= ∠4$.
在$\triangle ABC和\triangle CDA$中,
$\left\{\begin{array}{l} ∠1= ∠2,\\ AC= CA(公共边),\\ ∠4= ∠3,\end{array} \right. $
$\therefore \triangle ABC\cong \triangle CDA$().

【分析】解题的关键是由两组平行线得出两组角分别相等,根据公共边构造两角及其夹边分别相等.
证明:$\because AB// DC$,$AD// BC$,
$\therefore ∠1= ∠2$,$∠3= ∠4$.
在$\triangle ABC和\triangle CDA$中,
$\left\{\begin{array}{l} ∠1= ∠2,\\ AC= CA(公共边),\\ ∠4= ∠3,\end{array} \right. $
$\therefore \triangle ABC\cong \triangle CDA(ASA)$.

【例2】如图,$AB= AC$,$AD= AE$,$BE与CD相交于点O$.求证$\triangle BOD\cong \triangle COE$.



证明:在$\triangle ABE和\triangle ACD$中,
$\left\{\begin{array}{l} AB= AC,\\ ∠A= ∠A(公共角),\\ AE= AD,\end{array} \right.$
$\therefore \triangle ABE\cong \triangle ACD$().
$\therefore ∠B= ∠C$.
$\because AB= AC$,$AD= AE$,
$\therefore AB-AD= AC-AE$,即.
在$\triangle BOD和\triangle COE$中,
$\left\{\begin{array}{l} ∠B= ∠C,\\ ∠BOD= ∠COE,\\ BD= CE,\end{array} \right.$
$\therefore \triangle BOD\cong \triangle COE$().

【分析】找出两个三角形中两个角及其中一角的对边分别相等,利用“AAS”判定两个三角形全等.
证明:在$\triangle ABE和\triangle ACD$中,
$\left\{\begin{array}{l} AB= AC,\\ ∠A= ∠A(公共角),\\ AE= AD,\end{array} \right. $
$\therefore \triangle ABE\cong \triangle ACD(SAS)$.
$\therefore ∠B= ∠C$.
$\because AB= AC$,$AD= AE$,
$\therefore AB-AD= AC-AE$,即$BD= CE$.
在$\triangle BOD和\triangle COE$中,
$\left\{\begin{array}{l} ∠B= ∠C,\\ ∠BOD= ∠COE,\\ BD= CE,\end{array} \right. $
$\therefore \triangle BOD\cong \triangle COE(AAS)$.

一、选择题
1. 如图,玻璃三角板摔成三块.现在到玻璃店再配一块同样大小的三角板,最省事的方法是()

A.带①去
B.带②去
C.带③去
D.带①②③去

1. C

2. 如图,$∠1= ∠2$,则添加下列条件不一定能使$\triangle ABD\cong \triangle ACD$的是()


A.$AB= AC$
B.$BD= CD$
C.$∠B= ∠C$
D.$∠BDA= ∠CDA$

2. B