The triangles ABC and DEF have AD, BE and CF parallel. Show that [AEF] + [DBF] + [DEC] + [DBC] + [AEC] + [ABF] = 3 [ABC] + 3 [DEF], where [XYZ] denotes the signed area of the triangle XYZ. Thus [XYZ] is + area XYZ if the order X, Y, Z is anti-clockwise and - area XYZ if the order X, Y, Z is clockwise. So, in particular, [XYZ] = [YZX] = -[YXZ].
Solution
The starting point is that [ABC] = [XBC] + [AXC] + [ABX] (*) for any point X. If X is inside the triangle, then all the rotations have the same sense. If X is outside, then they do not. But it is easy to check that (*) always holds.
So [ABC] = [DBC] + [ADC] + [ABD], [DEF] = [AEF] + [DAF] + [DEA]. Now, ignoring sign, the triangles ABD and DEA have equal area, because they have a common base AD and the same height (since AD is parallel to BE). But the sign is opposite, so [ABD] + [DEA] = 0. Similarly, [ADC] + [DAF] = 0, so [ABC] + [DEF] = [DBC] + [AEF]. Adding the two similar equations (obtained from E with [ABC] and B with [DEF], and from F with [ABC] and C with [DEF]) gives the required result.
Alternative solution
Use vectors. Take origin A. Put b = AB, c = AC, d = AD. Then AE = b + h d, AF = c + k d, for some real h, k. Then 2 [ABC] = b x c etc and the rest is purely mechanical. Start with the lhs, expand and use d x d = 0.
© John Scholes
jscholes@kalva.demon.co.uk
20 Aug 2002