In triangle ABC, the medians AD, BE, and CF concur at the centroid G.
Prove that AD<(AB+AC)/2.
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
Yes Alan thanks. That is a really simple way to demonstrate that this relationship must be true and I am also impressed.
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
<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
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?
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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