Find the Ratio of the area of the triangle DEC to the Area of the parallelogram ABCD
When: AF is 1 FB is 2 BC is 2
Triangle DEC is similar to triangle BEF, (angle DEC = angle BEF as they are
vertical angles, and angle EDC = angle FBE)
So take DC and BF as the bases of these two triagles, they are in the ratio
of 3:2 (given). Hence their heights are in the ratio 3:2, so the heights of
triangle DEC and parallologram ABCD are in the ratio of 3:5
Hence as the area of a triangle is base*height/2 and that of a
parallellogram is base times height, and they share a common base the areas
of the triangle and parallelogram are in the ratio 3/2:5 or 3:10.
RonL