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# Help On Geometric Proof

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

Guest Feb 21, 2017

#1
+26366
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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
Sort:

#1
+26366
+10

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
#2
+79881
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Very nice, Alan.... genius in simplicity....!!!!!

CPhill  Feb 23, 2017
#3
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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

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.

Would this pass as a formal proof?

Melody  Feb 24, 2017
#4
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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.

This is a major reason why it is so much better to be a member.

If you are a member you can get email notifications that an answer has come in.