4.7 SAS三角和谐

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  SAS 三角和谐

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  Side-Angle-Side Postulate
  ::侧边角- 侧角- 侧边侧侧侧侧侧侧侧侧

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  If two sides and the included angle in one triangle are congruent to two sides and the included angle in another triangle, then the two triangles are congruent. (When an angle is between two given sides of a polygon it is called an included angle .)
  ::如果两边和一个三角形中包含角与两边一致,而另一三角形中包含角相同,则两个三角形是相同的。 (当一个角在多边形的两个指定边之间时,它被称为一个包含角。 )

  

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  $\begin{array}{r}\overline{AC}\cong \overline{XZ},\overline{BC}\cong \overline{YZ}\end{array}$ , and $\begin{array}{r}\mathrm{\angle }C\cong \mathrm{\angle }Z\end{array}$ , then $\begin{array}{r}\mathrm{△}ABC\cong \mathrm{△}XYZ\end{array}$ .
  ::ACXZ,BCYZ,CC,然后ABCXYZ。

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  This is called the Side-Angle-Side (SAS) Postulate and it is a shortcut for proving that two triangles are congruent. The placement of the word Angle is important because it indicates that the angle you are given is between the two sides.
  ::这被称为侧角- 角- 侧侧( SAS) 假设, 用来证明两个三角是相似的快捷方式 。 安格( angle) 字的位置很重要 , 因为它表明您所给的角是介于两边的 。

  

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  $\begin{array}{r}\mathrm{\angle }B\end{array}$ would be the included angle for sides $\begin{array}{r}\overline{AB}\end{array}$ and $\begin{array}{r}\overline{BC}\end{array}$ .
  ::B是AB和BC两边的角

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  What if you were given two triangles and provided with only two of their side lengths and the measure of the angle between those two sides? How could you determine if the two triangles were congruent?
  ::假若你们获得两个三角形,而你们得享受其中两个三角形的侧边长度和两边角的尺度,你们怎么确定这两个三角形是否一致呢?

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  Examples
  ::实例

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  Example 1
  ::例1

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  Is the pair of triangles congruent? If so, write the congruence statement and why.
  ::三角形是否一致?如果一致,请写出一致声明和原因。

  

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  The pair of triangles is congruent by the postulate. $\begin{array}{r}\mathrm{△}CAB\cong \mathrm{△}QRS\end{array}$ .
  ::三角形的对数与假设值相同 。 @CABRS 。

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  Example 2
  ::例2

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  State the additional piece of information needed to show that each pair of triangles is congruent.
  ::显示每对三角形的相近性 所需的附加信息 。

  

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  We know that one pair of sides and one pair of angles are congruent from the diagram . In order to know that the triangles are congruent by SAS we need to know that the pair of sides on the other side of the angle are congruent. So, we need to know that $\begin{array}{r}\overline{EF}\cong \overline{BA}\end{array}$ .
  ::我们知道一对一对和一对角度与图表是相同的。为了知道三角形与SAS是相同的,我们需要知道角的另一侧的两对是相同的。因此,我们需要知道EF _BA。

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  Example 3
  ::例3

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  Fill in the blanks in the proof below. 
  ::填充以下证据中的空白。

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  Given :
  ::参照:

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  $\begin{array}{r}\overline{AE}\cong \overline{DE},\overline{BE}\cong \overline{CE}\end{array}$
  ::AEDE,是的,是的

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  Prove : $\begin{array}{r}\mathrm{△}ABE\cong \mathrm{△}ACE\end{array}$
  ::证明:

  

  Statement	Reason
  1. $\begin{array}{r}\overline{AE}\cong \overline{DE},\overline{BE}\cong \overline{CE}\end{array}$	1. Given
  2. $\begin{array}{r}\mathrm{\angle }AEB\cong \mathrm{\angle }DEC\end{array}$	2. Vertical Angle Theorem
  3. $\begin{array}{r}\mathrm{△}ABE\cong \mathrm{△}ACE\end{array}$	3. SAS postulate

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  Example 4
  ::例4

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  What additional piece of information do you need to show that these two triangles are congruent using the SAS Postulate,  $\begin{array}{r}\mathrm{\angle }ABC\cong \mathrm{\angle }LKM\end{array}$ ,  $\begin{array}{r}\overline{AB}\cong \overline{LK}\end{array}$ ,  $\begin{array}{r}\overline{BC}\cong \overline{KM}\end{array}$ , or  $\begin{array}{r}\mathrm{\angle }BAC\cong \mathrm{\angle }KLM\end{array}$ ?
  ::您还需要什么信息来显示 使用SAS Postate, ABCLKM, AB LKK', BC KM', 或BACKLM, 这两个三角形是相同的?

  

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  For the SAS Postulate, you need the side on the other side of the angle. In $\begin{array}{r}\mathrm{△}ABC\end{array}$ , that is $\begin{array}{r}\overline{BC}\end{array}$ and in $\begin{array}{r}\mathrm{△}LKM\end{array}$ that is $\begin{array}{r}\overline{KM}\end{array}$ . The answer is $\begin{array}{r}\overline{BC}\cong \overline{KM}\end{array}$ .
  ::对于 SAS 假设, 您需要角度的另一侧 。 在 ABC 中, 是 BC , 而在 LKM 中是 KM 。 答案是 BC QKM 。

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  Example 5
  ::例5

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  Is the pair of triangles congruent? If so, write the congruence statement and why.
  ::三角形是否一致?如果一致,请写出一致声明和原因。

  

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  While the triangles have two pairs of sides and one pair of angles that are congruent, the angle is not in the same place in both triangles. The first triangle fits with SAS, but the second triangle is SSA. There is not enough information for us to know whether or not these triangles are congruent.
  ::虽然三角形有两对侧面和一对角度,但角在两个三角形中的位置不同。 第一个三角形符合SAS, 但第二个三角形是SSA。没有足够的信息让我们知道这些三角形是否一致。

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  Review
  ::回顾

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  Are the pairs of triangles necessarily congruent? If so, write the congruence statement and why.
  ::两对三角形是否必然一致? 如果一致,请写出一致声明和原因。

  1. 
  2. 
  3. 

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  State the additional piece of information needed to show that each pair of triangles is congruent by SAS.
  ::显示每对三角形与SAS一致的附加信息 。

  4. 
  5. 
  6. 

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  Fill in the blanks in the proofs below.
  ::填充以下证据中的空白。

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  7. Given :
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     • $\begin{array}{r}B\end{array}$ is a midpoint of $\begin{array}{r}\overline{DC}\end{array}$
       ::B 是DC的中点
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     • $\begin{array}{r}\overline{AB}\perp \overline{DC}\end{array}$
       ::AB DC 的 AB DC
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     Prove : $\begin{array}{r}\mathrm{△}ABD\cong \mathrm{△}ABC\end{array}$

     
     
     ::证明:
     
     ::B是DC AB DC proove: ABD ABC的中点

  Statement	Reason
  1. $\begin{array}{r}B\end{array}$ is a midpoint of $\begin{array}{r}\overline{DC},\overline{AB}\perp \overline{DC}\end{array}$	1.
  2.	2. Definition of a midpoint
  3. $\begin{array}{r}\mathrm{\angle }ABD\end{array}$ and $\begin{array}{r}\mathrm{\angle }ABC\end{array}$ are right angles	3.
  4.	4. All right angles are $\begin{array}{r}\cong \end{array}$
  5.	5.
  6. $\begin{array}{r}\mathrm{△}ABD\cong \mathrm{△}ABC\end{array}$	6.

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  8. Given :
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     • $\begin{array}{r}\overline{AB}\end{array}$ is an angle bisector of $\begin{array}{r}\mathrm{\angle }DAC\end{array}$
       ::AB是ZZDAC的双角区块
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     • $\begin{array}{r}\overline{AD}\cong \overline{AC}\end{array}$
       ::AD AC AD AD AC AD AD AD AC AD AD AA AC AD AD AA AC AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA AA A AAAAAAAAA AA AA AAAA AAAAA AA AA AA AA AA AAAAA AAAAAAA AAAAAAAAAA A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A A A A A A A AAA A AAAAAAAAAAAAAAAAAAAAAAAAAA AAAA A AAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
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     Prove : $\begin{array}{r}\mathrm{△}ABD\cong \mathrm{△}ABC\end{array}$
     ::证明:

     
     ::参考:AB是ADAAD AD AC' Prove: ABD ABC的一个角度,

  Statement	Reason
  1.	1.
  2. $\begin{array}{r}\mathrm{\angle }DAB\cong \mathrm{\angle }BAC\end{array}$	2.
  3.	3. Reflexive PoC
  4. $\begin{array}{r}\mathrm{△}ABD\cong \mathrm{△}ABC\end{array}$	4.

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  9. Given :
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     • $\begin{array}{r}B\end{array}$ is the midpoint of $\begin{array}{r}\overline{DE}\end{array}$ and $\begin{array}{r}\overline{AC}\end{array}$
       ::B 是 DE 和 AC 的中点
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     • $\begin{array}{r}\mathrm{\angle }ABE\end{array}$ is a right angle
       ::ABE 是一个右角度
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     Prove : $\begin{array}{r}\mathrm{△}ABE\cong \mathrm{△}CBD\end{array}$
     ::证明: ABE-CBD

     
     ::说明:B是DE的中点,AC是正确角度,证明:`ABE'CBD

  Statement	Reason
  1.	1. Given
  2. $\begin{array}{r}\overline{DB}\cong \overline{BE},\overline{AB}\cong \overline{BC}\end{array}$	2.
  3.	3. Definition of a Right Angle
  4.	4. Vertical Angle Theorem
  5. $\begin{array}{r}\mathrm{△}ABE\cong \mathrm{△}CBD\end{array}$	5.

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  10. Given :
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      • $\begin{array}{r}\overline{DB}\end{array}$ is the angle bisector of $\begin{array}{r}\mathrm{\angle }ADC\end{array}$
        ::DB是 QADC 的侧角
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      • $\begin{array}{r}\overline{AD}\cong \overline{DC}\end{array}$
        ::ADD'DD'AD$ $AD$ $AD$ $AD$ $AD$ $AD$ $AD$ $AD$ $AD$ $ADD$$$ $AAD$$ $ADDD$ $AAD$$ $AAAD$$$$$$$$ $$$$ $$ $$ $$$ $$ $ $ $$$$$$$$$$$ $$$$$$$$$$$$$$$$$$$$$$$$$$$ $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ $ $ $$$ $ $ $
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      Prove : $\begin{array}{r}\mathrm{△}ABD\cong \mathrm{△}CBD\end{array}$
      ::证明:@ABD_CBD

      
      ::参考:DB是“ADC AD DC' Prove:'ABD'CBD”的分角。

  Statement	Reason
  1.	1.
  2. $\begin{array}{r}\mathrm{\angle }ADB\cong \mathrm{\angle }BDC\end{array}$	2.
  3.	3.
  4. $\begin{array}{r}\mathrm{△}ABD\cong \mathrm{△}CBD\end{array}$	4.

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。

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  Resources
  ::资源