Prove that for every two triangles, there exists a line that halves the areas of the triangles simultaneously. The problem may sound somewhat funny, cause I haven't got it from a good source. Anyways, I think I somehow get what it wants and here's my way of looking at the solution:
Two triangles of arbitrary sizes are given. Let's halve (the area of) one of them with a line. (Can we do that, at all?) Then we slide this triangle to cut the line and go through it like it does in the figure. Take the area above this line $A$. Then $A$ is 0 in the beginning and will eventually be the area of the whole triangle, $A_1$. i.e. $0\leq A\leq A_1$. As the triangle goes on sliding, from the intermediate value theorem, there is a position at which the same line halves the triangle's area. (I'm not sure whether it's right to say this!)

I can go on with a function and say there exists a value on which the function gives us the half of the triangle on one side and the other half on the other. I just want to check out whether my approach is correct or not? Any response is appreciated.

Thanks to @Ihf, and Computational Geometry, by Ham Sandwich Theorem the problem is solved!
Here's the Ham Sandwich Theorem: 
In mathematical measure theory, the sandwich theorem states that given $n$ measurable "objects" in $n$-dimensional space, it is possible to divide all of them in half (with respect to their measure, i.e. volume) with a single $(n − 1)$-dimensional hyperplane.
In case of our problem, take the objects 2 triangles in the 2-D  space (plane). Now it's possible to divide these two triangle in half (their areas) with a single 1-D hyperplane (a line).
That's it!
