The three corner triangles have the same area because if you take any two of them their bases and heights are in inverse proportion.
Area(ΔAFD)=Area(ΔBDE)=Area(ΔCEF)
So
Area(ΔDEF)=Area(ΔABC)3*Area(ΔAFD)
Let AC=b be the base of ΔABC and the height=h, area = bh/2
The base and height of ΔAFD are then (1x)b and xh, area = x(1x)bh/2
Area(ΔDEF) = bh/2  3x(1x)bh/2
The ratio sought is then 1  3x(1x) = 3x^{2}  3x + 1

Posted by Jer
on 20171126 11:16:42 