Where would this segment DE be?
By: Tim Lehman

For the diagram below, we want to find points D and E so that segment DE is parallel to side AC and divides triangle ABC into two equal pieces. Because DE is parallel to AC, triangles BDE and BAC are similar. For the area of triangle BDE to be equal to half of the area of triangle BAC, the scale factor of the sides of BAC to BDE must be 1 to the square root of 2.

Thus, the ratio of BA to BD is 1 to square root of 2. Because the ratio of the legs to the hypotenuse of a 45-45-90 triangle is 1 to the square root of 2, we only need to construct a 45-45-90 triangle with leg length the same as BD and the length of the hypotenuse will be the same as BA.

Thus, we begin by constructing point F such that angle ABF = angle FAB = 45 degrees. (If general ruler and compass rules apply, simply construct a line passing through points A and B and perpendicular to AB and then bisect these right angles.) By construction, triangle AFB is a 45-45-90 triangle. Therefore, point D lies on the circle centered at B and passing through F. Point E can then be found by constructing the line parallel to segment AC and passing through point D.