Math Is Fun Forum

Maroon900

Member

Registered: 2017-02-21

Posts: 1

Geometry Proof.

Hello! I dont get this. I tried adding the inequalities but I dont know which to add. I really appreciate if you could help me with part (a).

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

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

Bob

Administrator

Registered: 2010-06-20

Posts: 10,640

Re: Geometry Proof.

hi Maroon900

Welcome to the forum.

At first I thought this was a use of the triangle inequality and, after a bit of manipulating, I got AD < AB + AC.  Nearly there I thought.  Then I tried a fresh approach:

Rotate the triangle about point D to create a new triangle A'BC below the first.  The resulting shape ABA'C is a parallelogram (can you prove this?) and the diagonal is 2AD.  You'll get the required result from this.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Freiza

Member

Registered: 2017-02-21

Posts: 6

Re: Geometry Proof.

hi Maroon900

Welcome to the forum.

At first I thought this was a use of the triangle inequality and, after a bit of manipulating, I got AD < AB + AC.  Nearly there I thought.  Then I tried a fresh approach:

Rotate the triangle about point D to create a new triangle A'BC below the first.  The resulting shape ABA'C is a parallelogram (can you prove this?) and the diagonal is 2AD.  You'll get the required result from this.

Bob

Brilliant. Creating parallelogram was the trick.

Maroon900, Hint, SSS and 2AD < AC+ A'C

Last edited by Freiza (2017-02-21 21:33:17)