To trisect the area of a triangle ABC, construct the square root of "6" times BC. This is found by the spiral triangle construction at the bottom. CG is equal to root "6" times BC. Click here for a full explaination of this square root construction. Trisect this segment CG and use that radius and center C to draw a circle. Where the circle intersects segement BC, put a point (not labelled). Draw a line segement through this point parallel to line segment CA intersecting line segment BA at a point. Call this point D. Now draw a line parallel to BC through D intersecting line segment AC at point E. Line segment DE trisects the area of triangle ABC.

Click here to return to Ian's Geometry Forum.