ABCD is a square of side length 1. E is on AB and F is on BC such that DE and DF trisect angle ADC. Find the area of triangle DEF.
Since ABCD is a square, each of its corner angles are right angles, each = 90o.
Since DE and DF trisect angle(ADC), angle(ADE) = 30o, angle(EDF) = 30o, angle(FDC) = 30o.
Triangle(DCF) is a 30o - 60o - 90o right triangle.
Since DC = 1, CF = 1/sqrt(3) and DF = 2/sqrt(3).
Similarly, DE = 2/sqrt(3).
Using this formula to find the area of triangle(EDF): area = ½·a·b·sin(C)
---> area = ½ · 2/sqrt(3) · 2/sqrt(3) ·sin(30o)
---> area = ½ · 2/sqrt(3) · 2/sqrt(3) · 0.50
---> area = 1/3