USAMO 1977 Problem 2

The triangles ABC and DEF have AD, BE and CF parallel. Show that $[\mathrm{AEF}]+[\mathrm{DBF}]+[\mathrm{DEC}]+[\mathrm{DBC}]+$ $[\mathrm{AEC}]+[\mathrm{ABF}]=3[\mathrm{ABC}]+3[\mathrm{DEF}]$, where $[\mathrm{XYZ}]$ denotes the signed area of the triangle XYZ. Thus $[\mathrm{XYZ}]$ is + area XYZ if the order $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ is anti-clockwise and - area $\mathrm{XYZ}$ if the order $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ is clockwise. So, in particular, $[\mathrm{XYZ}]=[\mathrm{YZX}]=-[\mathrm{YXZ}]$.