We know that the line DE is parallel to the line BC. Give that the area of triangle ADE is 9 and the area of triangle CDE is 6, find the area of Atriangle BC 
+1694 Angle A = 45º
AD = DE = 3√2
CD = 12 / 3√2 = 2√2
[BCE] = 6 + (2√2)2 / 2
[ABC] = [ADE] + [CDE] + [BCE]
