December 6th, 2010, 08:03 PM  #1 
Member Joined: Jun 2010 Posts: 44 Thanks: 0  triangle geometry
O is any point inside the triangle ABC. AO, BO, CO are joined and produced to meet the opposite sides BC, CA, and AB at D, E and F respectively. Prove that AO/AD+BO/BE+CO/CF=2 Obviously, O is meant to be the centroid. The question did not say that the lines produced cut the opposite sides into two equal lengths, where D, E and F are midpoints of the respective sides they cut. Shouldn't the question include this piece of information? 
December 6th, 2010, 08:36 PM  #2 
Member Joined: Jul 2009 From: New Jersey Posts: 65 Thanks: 0  Re: triangle geometry
Why does it necessarily have to be the centroid?

December 7th, 2010, 08:41 AM  #3  
Member Joined: Jun 2010 Posts: 44 Thanks: 0  Re: triangle geometry Quote:
 
December 7th, 2010, 02:51 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,271 Thanks: 1959 
The lines AO, BO and CO divide triangle ABC into various triangles, whose areas are related to each other in certain fairly obvious ways. The equation you have to prove can be written as a relationship between those areas, and can be shown to follow from the relationships I mentioned previously.

December 8th, 2010, 07:46 AM  #5  
Member Joined: Jun 2010 Posts: 44 Thanks: 0  Re: Quote:
 
December 8th, 2010, 11:29 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,271 Thanks: 1959 
Most of them. You might be able to avoid explicitly mentioning a very few. Various pairs of triangles have an altitude in common, making it worthwhile to consider the ratio of their areas. You can get a good idea of the type of proof you need to find by looking up a proof of Ceva's theorem, and you'll need to come within a whisker of proving that theorem as part of solving this problem.

December 8th, 2010, 08:37 PM  #7  
Member Joined: Jun 2010 Posts: 44 Thanks: 0  Re: Quote:
 

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