A geometric problem almost solved

• Jun 14th 2011, 01:11 AM
Sarasij
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)...

• Jun 14th 2011, 01:13 AM
Sarasij
Re: A geometric problem almost solved
Quote:

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)...