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)...
Please help me here guys ...
I am not sure how to constructively find the required line, but there is a more general theorem that any finite shape on a plane can be divided into equal areas by a line parallel to a given line. It uses the Intermediate value theorem from calculus, but this theorem is very intuitive. This is also discussed in the book "Penrose tiles to trapdoor ciphers" by Martin Gardner. Unfortunately, its Google preview lacks the key page. Nevertheless, you can start reading on p. 167 from "Let us drop the requirement of congruence...". The idea is that we take a line parallel to AO that is located above the triangle, so that the part of the triangle above the line is 0%. We then move the line down until 100% of the triangle is above the line. There has to be a point where exactly 50% of the triangle is above the line.