+0

# Help On Geometric Proof

+10
429
4

In triangle ABC, the medians AD, BE, and CF concur at the centroid G. Feb 21, 2017

#1
+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

.

Feb 23, 2017

#1
+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
0

Very nice, Alan.... genius in simplicity....!!!!!   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

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?

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.

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.

Your question also gets automatically stored in your watch list  and you can see notification there if you have a new answer.

Feb 24, 2017