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avatar+2441 
+1

1) I am fairly certain that you are referring to the midsegment theorem. Here is a diagram for you to reference as I prove:

 


 

We know that by the midsegment theorem that the segment bisects both of its intersecting sides and is parallel to the third side. \(\angle CDE\cong\angle CAB\) and \(\angle CED\cong\angle CBA\) by the Corresponding Angles Postulate. By the Angle-Angle Similarity Theorem, \(\triangle CDE \sim \triangle CAB\). We know that \(\frac{CD}{CA}\) equals some ratio. We know that \(\overline{CA}\) is broken up into two smaller segments, \(\overline{CD}\) and \(\overline{DA}\). The ratio is now \(\frac{CD}{CD+DA}\). By the given info, \(CD=DA\), so, by the substitution property of equality, the ratio can be written as\(\frac{CD}{CD+CD}=\frac{CD}{2CD}=\frac{1}{2}\). Since the ratio of corresponding sides of a similar triangle are equal, the ratio of \(\frac{DE}{AB}=\frac{1}{2}\). Rewriting this ratio turns into \(2DE=AB\Rightarrow DE=\frac{1}{2}AB\).

 

2) Here is a diagram again!
 

 

In this diagram here, \(AB=AC\), and \(D\) is the midpoint of \(\overline{AB}\) and \(E\) is the midpoint of \(\overline{AC}\), which makes \(\overline{BE}\) and \(\overline{CD}\) medians of this isosceles triangle.

 

By the given info, \(AB=AC\) and \(AD=AE\) because they are midpoints of equal-length segments. \(\angle A\cong\angle A\) by the reflexive property of congruence. Therefore, by Side-Angle-Side Congruency Theorem, \(\triangle ABE\cong\triangle ACD\). Utilizing the fact that corresponding parts of congruent triangles are congruent, \(CD=BE\). We are done!

Nov 26, 2017
 #3
avatar+2441 
+1

Of course! I will gladly delve deeper into this subject.

 

The explanation I provided is probably unsatisfactory and shoddy anyway. I believe that these rules are best demonstrated by example.

 

1. Divisibility by 7

 

Is \(205226\) divisible by 7? Well, let's use the process!

 

\(\textcolor{blue}{20522}\textcolor{red}{6}\)  
1. \(\textcolor{blue}{20522}-2*\textcolor{red}{6}=\textcolor{blue}{2051}\textcolor{red}{0}\) I have no clue if this is indeed divisible by 7, so do this process again and again (hence recursion)

2. 

\(\textcolor{blue}{2051}-2*\textcolor{red}{0}=\textcolor{blue}{205}\textcolor{red}{1}\)

I still cannot tell, so I will do this again.

3.

\(\textcolor{blue}{205}-2*\textcolor{red}{1}=\textcolor{blue}{20}\textcolor{red}{3}\)

I still cannot tell.

4.

\(\textcolor{blue}{20}-2*\textcolor{red}{3}=\textcolor{blue}{1}\textcolor{red}{4}\)

I know that 14 is divisible by 7, so the original number is, too.

 

How about \(22604\)? Well, let's check it!

 

\(\textcolor{blue}{2260}\textcolor{red}{4}\)  

1.

\(\textcolor{blue}{2260}-2*\textcolor{red}{4}=\textcolor{blue}{225}\textcolor{red}{2}\)

Yet again, I cannot make a judgment.

2.

\(\textcolor{blue}{225}-2*\textcolor{red}{2}=\textcolor{blue}{22}\textcolor{red}{1}\)

Of course, we must keep going.

3.

\(\textcolor{blue}{22}-2*\textcolor{red}{1}=20\)

I know that this number is not divisible by 7, so the original number is not either.
   

 

2. Divisibility by 11

 

Let's check if \(43923\) is divisible.

 

\(\textcolor{blue}{4392}\textcolor{red}{3}\)  

1.

\(\textcolor{blue}{4392}-\textcolor{red}{3}=\textcolor{blue}{438}\textcolor{red}{9}\)

Let's do it again!

2.

\(\textcolor{blue}{438}-\textcolor{red}{9}=\textcolor{blue}{42}\textcolor{red}{9}\)

One more time!
\(\textcolor{blue}{42}-\textcolor{red}{9}=\textcolor{blue}{3}\textcolor{red}{3}\) I know that 33 is divisible by 11, so the original number is, too.
   

 

How about \(123567\)?

 

\(\textcolor{blue}{12356}\textcolor{red}{7}\)

 

1.

\(\textcolor{blue}{12356}-\textcolor{red}{7}=\textcolor{blue}{1234}\textcolor{red}{9}\)

This requires perserverance. Keep going!

2.

\(\textcolor{blue}{1234}-\textcolor{red}{9}=\textcolor{blue}{122}\textcolor{red}{5}\)

 

3.

\(\textcolor{blue}{122}-\textcolor{red}{5}=\textcolor{blue}{11}\textcolor{red}{7}\)

I know that \(11*11=121\), so 117 is not divisible.
   

 

3. Divisibility by 13

 

Is \(19704\) divisible? Let's find out!

 

\(\textcolor{blue}{1970}\textcolor{red}{4}\)  

1.

\(\textcolor{blue}{1970}+4*\textcolor{red}{4}=\textcolor{blue}{198}\textcolor{red}{6}\)

 

2.

\(\textcolor{blue}{198}+4*\textcolor{red}{6}=\textcolor{blue}{22}\textcolor{red}{2}\)

 
\(=\textcolor{blue}{22}+4*\textcolor{red}{2}=\textcolor{blue}{3}\textcolor{red}{0}\) \(13*3=36\), so 30 is not divisible by 13 and nor is the given number.
   

 

Is \(9321 \) able to be divised?
 

\(\textcolor{blue}{932}\textcolor{red}{1}\)  

1.

\(\textcolor{blue}{932}+4*\textcolor{red}{1}=\textcolor{blue}{93}\textcolor{red}{6}\)

 

2.

\(\textcolor{blue}{93}+4*\textcolor{red}{6}=\textcolor{blue}{11}\textcolor{red}{7}\)

 

3.

\(\textcolor{blue}{11}+4*\textcolor{red}{7}=\textcolor{blue}{3}\textcolor{red}{9}\)

39 is divisible by 13, so the original number is as well.
   
Nov 26, 2017
Nov 25, 2017
 #2
avatar+118608 
+2
Nov 25, 2017
 #7
avatar+33616 
+3
Nov 25, 2017

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