Some fascinating results motivated me to make this Demonstration. For instance, any line through the incenter of a triangle (where the three internal angle bisectors meet) that divides its area in half also divides its perimeter in half; conversely, any line through the incenter that divides the perimeter of the triangle in half also divides its area in half. More generally, we have Haider's theorem: for any triangle and any line , divides the area and perimeter of in the same ratio if and only if it passes through the incenter of .
One of the difficult problems designed to eliminate many targeted applicants in entrance examinations to the Mekh-Mat at Moscow State University during the 1970s and 1980s was to draw a straight line that bisects the area and perimeter of a triangle. Showing the existence of such a line is quite easy; constructing that line is much harder.