Here is a cute triangle result that I’m surprised I had not known previously. If we are given a point on one of the sides of a triangle, how do we find a line through the triangle that cuts its area in half?
Clearly if that point is either a midpoint or one of the vertices, the answer is a median of the triangle. A median cuts a triangle in half since the two pieces have the same length side and equal height.
So what if the point is not a midpoint or a vertex? Referring to the diagram below, if is our desired point closer to than , the end point of the area-bisecting segment would need to be on side so that area(BPQ) = area(ABC)/2.
In other words, we require area(BPQ) = area(BDQ), or, subtracting the areas of triangle BDQ from both sides,
Since these two triangles share the common base , this tells us that we require them to have the same height. In other words, we require to be parallel to . This tells us how to construct the point given on :
- Construct the midpoint D of .
- Draw parallel to .
See  for an animation of this construction.
In turns out that the set of all area-bisecting lines are tangent to three hyperbolas and enclose a deltoid of area times the original triangle. [2,3,4]
 Jaime Rangel-Mondragon, “Bisecting a Triangle” http://demonstrations.wolfram.com/BisectingATriangle/ from the Wolfram Demonstrations Project Published: July 10, 2013