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I'm not sure where to start with this problem:

 

In triangle ABC, the medians line AD, line BE, and line CF concur at the centroid G.

(a) Prove that AD < (AB + AC)/2.

 

(b) Let P = AB + AC + BC be the perimeter of triangle ABC. Prove that\( \frac{3P}{4} < AD + BE + CF < P.\)

 

Thanks so much!

 #1
avatar+92805 
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Hi,

I am not great at geometry questions but sill I am wondering....

Are you sure part a is correct?

If all those lengths are squared then I can prove it.

Melody  Apr 9, 2018
 #2
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Part a is correct it was directly copied from the problem I was presented with in my book :)

 #3
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im not sure bro

jakesplace  Apr 9, 2018
 #4
avatar+428 
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im not the geometry wizard

jakesplace  Apr 9, 2018
 #5
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The concept that you need to use is the triangle inequality:

the sum of the lengths of two sides of a triangle will be greater than the length of the third side.

So, for example, from your diagram,

\(\displaystyle AG + GB > AB.\)

 

For part (a), extend the line AD to AA' such that ABA'C is a parallelogram and then think of how to make use of the long diagonal AA'.

 

For  the right hand part of the inequality in part (b) you need to make use of the result from part (a) together with the the other two possible inequalities of this type.

For the left hand part, it's back to the triangle inequality, and the clue is the inequality that I've stated on the fourth line, above. You will also have to make use of the 2 : 1 property for the medians of a triangle.

 

Tiggsy

Guest Apr 9, 2018

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