# Thread: A geometric problem almost solved

1. ## A geometric problem almost solved

ABC is any triangle. AO is any line segment. Prove that there exists a line segment parallel to AO which divides ABC into 2 equal areas.

Solution: I have made a construction to solve this. See the attached picture,there is a median AP from A,and I have chosen such a line DE which cuts AP at X in such a way that DE||AO and area(AXD)=area(PXE)[though its not clear from the badly drawn picture].

My question is that how can I prove the existence of such same area triangles as in this figure...bcoz once this can be proved,its an easy proof to show that quadrilateral(ADEC) has the same area as triangle(BDE)...

2. ## Re: A geometric problem almost solved

Originally Posted by Sarasij
ABC is any triangle. AO is any line segment. Prove that there exists a line segment parallel to AO which divides ABC into 2 equal areas.

Solution: I have made a construction to solve this. See the attached picture,there is a median AP from A,and I have chosen such a line DE which cuts AP at X in such a way that DE||AO and area(AXD)=area(PXE)[though its not clear from the badly drawn picture].

My question is that how can I prove the existence of such same area triangles as in this figure...bcoz once this can be proved,its an easy proof to show that quadrilateral(ADEC) has the same area as triangle(BDE)...