The three corner triangles have the same area because if you take any two of them their bases and heights are in inverse proportion.
Let AC=b be the base of ΔABC and the height=h, area = bh/2
The base and height of ΔAFD are then (1-x)b and xh, area = x(1-x)bh/2
Area(ΔDEF) = bh/2 - 3x(1-x)bh/2
The ratio sought is then 1 - 3x(1-x) = 3x2 - 3x + 1
Posted by Jer
on 2017-11-26 11:16:42