1. 两角及其
夹边
分别相等的两个三角形全等(简写成“角边角”或“ASA
”).答案：夹边 ASA
2. 符号语言:
如图,在$\triangle ABC$和$\triangle DEF$中,
$\left\{\begin{array}{l} ∠A=
$\therefore \triangle ABC\cong \triangle DEF(
如图,在$\triangle ABC$和$\triangle DEF$中,
$\left\{\begin{array}{l} ∠A=
∠D
,\\ AB=DE
,\\ ∠B=∠E
, \end{array}\right.$$\therefore \triangle ABC\cong \triangle DEF(
ASA
).$答案：$ \angle D $ $ DE $ $ \angle B = \angle E $ ASA
1. 如图,O是AB的中点,$∠A=∠B$. 求证:$\triangle AOC\cong \triangle BOD.$
答案：证明: $ \because O $ 是 $ AB $ 的中点, $ \therefore AO = BO $.
在 $ \triangle AOC $ 和 $ \triangle BOD $ 中, $ \left\{ \begin{array} { l } { \angle A = \angle B, } \\ { AO = BO, } \\ { \angle AOC = \angle BOD, } \end{array} \right. $
$ \therefore \triangle AOC \cong \triangle BOD ( ASA ) $.
在 $ \triangle AOC $ 和 $ \triangle BOD $ 中, $ \left\{ \begin{array} { l } { \angle A = \angle B, } \\ { AO = BO, } \\ { \angle AOC = \angle BOD, } \end{array} \right. $
$ \therefore \triangle AOC \cong \triangle BOD ( ASA ) $.
2. 如图,点B,D,C,F在同一条直线上,$AC// EF,BC=EF,∠B=∠CPD$,AB与DE相等吗? 说说你的理由.
答案：解: $ AB = DE $. 理由如下: $ \because AC // EF $,
$ \therefore \angle ACB = \angle F, \angle CPD = \angle E $,
$ \because \angle B = \angle CPD, \therefore \angle B = \angle E $.
在 $ \triangle ACB $ 和 $ \triangle DFE $ 中,
$ \left\{ \begin{array} { l } { \angle ACB = \angle F, } \\ { BC = EF, } \\ { \angle B = \angle E, } \end{array} \right. $ $ \therefore \triangle ACB \cong \triangle DFE ( ASA ) $,
$ \therefore AB = DE $.
$ \therefore \angle ACB = \angle F, \angle CPD = \angle E $,
$ \because \angle B = \angle CPD, \therefore \angle B = \angle E $.
在 $ \triangle ACB $ 和 $ \triangle DFE $ 中,
$ \left\{ \begin{array} { l } { \angle ACB = \angle F, } \\ { BC = EF, } \\ { \angle B = \angle E, } \end{array} \right. $ $ \therefore \triangle ACB \cong \triangle DFE ( ASA ) $,
$ \therefore AB = DE $.
3. 如图,点E在AB上,$CE=CB$,EC是$∠BED$的平分线,$∠DCA=∠BCE$. 求证:$CD=CA.$
答案：证明: $ \because EC $ 平分 $ \angle BED, \therefore \angle CED = \angle CEB $.
$ \because CE = CB, \therefore \angle CEB = \angle B, \therefore \angle CED = \angle B $.
$ \because \angle DCA = \angle BCE, \therefore \angle DCE = \angle ACB $.
在 $ \triangle DCE $ 和 $ \triangle ACB $ 中, $ \left\{ \begin{array} { l } { \angle DCE = \angle ACB, } \\ { CE = CB, } \\ { \angle DEC = \angle B, } \end{array} \right. $
$ \therefore \triangle DCE \cong \triangle ACB ( ASA ) $,
$ \therefore CD = CA $.
$ \because CE = CB, \therefore \angle CEB = \angle B, \therefore \angle CED = \angle B $.
$ \because \angle DCA = \angle BCE, \therefore \angle DCE = \angle ACB $.
在 $ \triangle DCE $ 和 $ \triangle ACB $ 中, $ \left\{ \begin{array} { l } { \angle DCE = \angle ACB, } \\ { CE = CB, } \\ { \angle DEC = \angle B, } \end{array} \right. $
$ \therefore \triangle DCE \cong \triangle ACB ( ASA ) $,
$ \therefore CD = CA $.