Let ABCDEF be a convex hexagon whose opposite sides are parallel.

Prove that triangles ACE and BDF have the same area.

In 'Find the Area' (2011.02.09) I laboriously showed that:
"The diagonals of a trapezoid divide it into four triangles. Two of the triangles (those having a side in common with the lateral  sides of the trapezoid) have equal areas."

Bractals responded with a much neater proof of the same notion:
Let WXYZ be an trapezoid with side WX parallel with side YZ and diagonals WY and XZ intersecting at point P.
   [WYX] = [WZX]
because they both have the same base and their altitudes are equal. Therefore,
   [XPY] = [WYX] - [WPX] = [WZX] - [WPX] = [ZPW].

The current hexagon can be divided into trapezoids in 3 different ways: ABDE, whose diagonals cross at T,BCEF, whose diagonals cross at U, and ACDF, whose diagonals cross at V. Now ATE = BTD, AVC = DVF,and CUE = BUF. There can be a small central piece TUV which is common to both triangles ACE and BDF.

Since ATE+AVC+CUE+TUV= ACE, and BTD+DVF+BUF+TUV=BDF, it follows immediately that ACE and BDF have the same area.

Edited on April 8, 2011, 2:35 am

Posted by broll on 2011-04-08 02:14:16