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

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

Guest Feb 21, 2017

#1**+10 **

Best Answer

Cut the triangle in two and rotate one part to make a new triangle as below (I've changed the notation somewhat):

Now we must have 2m < u + v or m < (u + v)/2

or AD < (AB + AC)/2

.

Alan Feb 23, 2017

#3**0 **

Yes Alan thanks. That is a really simple way to demonstrate that this relationship must be true and I am also impressed.

But

I am trying to work out how this could be proven in a formal proof......

I suppose you could write it exactly as you have done.

--------------------------

Consider the triangles ADB and ACD

AD bisects BC by definition of a median

therefore BD=DC

<BDA=180 - <ADC adjacent supplementary angles

Rotate triangle ADC 180 degrees about the point D. to from the new triangle AA'B

AC has been rotated to the position A'B

i.e AC=A'B

AA' < A'B+AB One side of any triangle must be less than the sum of the other 2 sides.

AD+DA' < A'B+AB

AD+AD < AC+AB

2AD < AC+AB

2AD < (AC+AB)/2

Would this pass as a formal proof?

Melody Feb 24, 2017

#4**0 **

**A note to all guests: :))**

I wonder if the asker of this question will ever see the answer.

The asker draw my attention to it with a repost. The question is 2 days old not.

I guess it is not likely to be seen, what a pity.

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Melody Feb 24, 2017