+0  
 
0
157
2
avatar+157 

In triangle ABC, the medians AD, BE, and 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 3P/4 < AD + BE + CF < P.

AdminMod2  Aug 15, 2017
Sort: 

1+0 Answers

 #1
avatar+78575 
+1

 

 

Since we can orient the triangle any way we want ...let  A  =(0,0) , B  = (0, b)  and C  = (c, 0)

 

 

Let  D  =   ( c/2 , b/2)    ...so  AD  =  sqrt [ (c/2 - 0)^2   +  (b /2 - 0) ^2  ]  =  sqrt [ ( b^2 + c^2 ) / 4 ]   = 

sqrt [ b^2 + c^2 ] / 2

 

AB  = b

 

And 

 

AC  = c

 

So

 

AD  <  [ AB  + AC] / 2

 

sqrt [ b^2 + c^2  ] / 2  <  [ b + c ] / 2

 

sqrt [ b^2 + c^2 ] < b + c        square both sides

 

b^2 + c^2  <  b^2  + 2bc + c^2

 

0 <  2bc       which is true  since  b, c   are both > 0 

 

 

 

cool cool cool

CPhill  Aug 15, 2017

6 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details