1.
两角
分别相等且其中一组等角的对边
相等的两个三角形全等(简写成“角角边”或“AAS
”).它是利用基本事实ASA得到的结论.答案：两角 对边 AAS
如图,在△ABC和△DEF中,$\left\{ \begin{array} { l } { \angle C = \_\_\_\_\_\_, } \\ { \_\_\_\_\_\_, } \\ { A B = D E, } \end{array} \right.$
∴△ABC≌△DEF(________).
∴△ABC≌△DEF(________).
∠F
∠A = ∠D
AAS
答案：$ \angle F $ $ \angle A = \angle D $ (或 $ \angle B = \angle E $) AAS
1. 如图,在△ABC和△ADE中,∠B=∠D,AC=AE,∠1=∠2.求证:BC=DE.
答案：证明: $ \because \angle 1 = \angle 2 $,
$ \therefore \angle DAC + \angle 1 = \angle 2 + \angle DAC $, 即 $ \angle BAC = \angle DAE $.
在 $ \triangle ABC $ 和 $ \triangle ADE $ 中, $ \left\{ \begin{array} { l } { \angle B = \angle D, } \\ { \angle BAC = \angle DAE, } \\ { AC = AE, } \end{array} \right. $
$ \therefore \triangle ABC \cong \triangle ADE ( AAS ) $,
$ \therefore BC = DE $.
$ \therefore \angle DAC + \angle 1 = \angle 2 + \angle DAC $, 即 $ \angle BAC = \angle DAE $.
在 $ \triangle ABC $ 和 $ \triangle ADE $ 中, $ \left\{ \begin{array} { l } { \angle B = \angle D, } \\ { \angle BAC = \angle DAE, } \\ { AC = AE, } \end{array} \right. $
$ \therefore \triangle ABC \cong \triangle ADE ( AAS ) $,
$ \therefore BC = DE $.
2. 如图,AB//CD,点E在CB的延长线上,∠A=∠E,AC=DE.求证:BC=CD.
答案：证明: $ \because AB // CD $, $ \therefore \angle ABC = \angle DCE $.
在 $ \triangle ABC $ 和 $ \triangle ECD $ 中,
$ \left\{ \begin{array} { l } { \angle A = \angle E, } \\ { \angle ABC = \angle DCE, } \\ { AC = DE, } \end{array} \right. $ $ \therefore \triangle ABC \cong \triangle ECD ( AAS ) $,
$ \therefore BC = CD $.
在 $ \triangle ABC $ 和 $ \triangle ECD $ 中,
$ \left\{ \begin{array} { l } { \angle A = \angle E, } \\ { \angle ABC = \angle DCE, } \\ { AC = DE, } \end{array} \right. $ $ \therefore \triangle ABC \cong \triangle ECD ( AAS ) $,
$ \therefore BC = CD $.
3. 如图,AC⊥CE,AB⊥BD,ED⊥BD,BC=DE.求证:△ABC≌△CDE.
答案：证明: $ \because AC \perp CE $, $ AB \perp BD $,
$ \therefore \angle A + \angle ACB = 90 ^ { \circ } $, $ \angle ACB + \angle ECD = 90 ^ { \circ } $,
$ \therefore \angle A = \angle ECD $.
$ \because AB \perp BD $, $ ED \perp BD $, $ \therefore \angle ABC = \angle CDE = 90 ^ { \circ } $.
在 $ \triangle ABC $ 和 $ \triangle CDE $ 中, $ \left\{ \begin{array} { l } { \angle A = \angle ECD, } \\ { \angle ABC = \angle CDE, } \\ { BC = DE, } \end{array} \right. $
$ \therefore \triangle ABC \cong \triangle CDE ( AAS ) $.
$ \therefore \angle A + \angle ACB = 90 ^ { \circ } $, $ \angle ACB + \angle ECD = 90 ^ { \circ } $,
$ \therefore \angle A = \angle ECD $.
$ \because AB \perp BD $, $ ED \perp BD $, $ \therefore \angle ABC = \angle CDE = 90 ^ { \circ } $.
在 $ \triangle ABC $ 和 $ \triangle CDE $ 中, $ \left\{ \begin{array} { l } { \angle A = \angle ECD, } \\ { \angle ABC = \angle CDE, } \\ { BC = DE, } \end{array} \right. $
$ \therefore \triangle ABC \cong \triangle CDE ( AAS ) $.