The above picture gives an example of a triangle whose inscribed rectangle
of maximum area has been constucted. Notice that the area of the rectangle
is half the area of the entire triangle. This will be the case for any inscribed
rectangle of maximum area.
The above picture will help to demonstate why the inscribed rectangle
of maximum area (the pink rectagle in this picture) has an area which is
half that of the big triangle. To make this picture, I rotated triangle
ADF 180 degrees about the point D to create triangle IDB. Similarly, triangle
BHE was created by rotating triangle EGC about point E. If we were to delete
triangles ADF and EGC, then what we would have left is two rectangles DIHE
and FDEG which have the same area. Thus, rectangle FDEG is 1/2 the big rectagle
FIHG. Since the big rectangle is equal in area to the big triangle that
we started with (because we simply rotated two parts of it to make the big
rectangle), rectangle FDEG = (triangle ABC)/2.
How do you know that rectangles DIHE and DFGE have equal area without using GSP to measure their areas? Let's look at the side where triangle ADF was rotated to make DIB. (A similar argument can be made for the other side with triangle BHE.) DIHE and DFGE have equal widths. ID = DF since DF was rotated to make ID. Both of these rectangles have a length that is equal to DE. Also angle AFD was right, so angle DIB is right. AD = DB since D is the midpoint of AB. When triangle ADF is rotated, AD "fits" exactly on DB.