Construct a triangle quadrisection?

Given triangle of sides a, b, c, we mark off points where the ratio of the short segments to the length of the sides are in the ratio of (1/2)(1 - 1/√5) = 0.276393... approximately. Then drawing lines from those points to the opposite vertex defines the interior triangle of area = 1/4 of original triangle. This ratio can be found by straightedge and compass, and I'll get to that later, as to a quick way to do it. Graphic to follow.

Edit: Let X, Y, Z be the vertices of a given triangle, and XY is the base. Draw line AB ⊥ XY passing through X, and mark a point C on XB such that AX = XC = 2XY, and CB = XY. Draw lines YA, YB. Then draw line DC II YB where D is on line YA, Mark E on line XY such that YE = YD. Find point F between EX such that EF = FX = (1/2)EX.

Next, mark point G on XY such that GY = XF. Draw line GH II YZ where H is on XZ, and draw line FI II YZ, where I is on line YZ. Mark point J on YZ such that JY = ZI. Then draw lines XJ, YH, ZF, which completes the drawing, after some erasing to make it look like the original diagram.

Graphic to come a bit later.

Edit 2: Here's the graphic:

http://i254.photobucket.com/albums/hh120/Scythian1...