The goal is to continue a perspective drawing of a floor tiled with congruent square tiles. Specifically, let XY be a horizontal line representing the horizon of the perspective drawing. Let ABCD be one of the square tiles in the foreground, with A nearest the viewer, B on the line XA, D on the line YA, and C the intersection of lines XD and YB. Show how to construct the tile next to ABCD with side BC in common with tile ABCD.

This is somewhat different from the previous methods, I think.

Let Z be the intersection of AC with XY and let W be the intersection of DB with XY.  Draw BZ and CW.  Let E be the intersection of BZ and CX and let F be the intersection of CW and BX.  Then BFEC is the tile sought.

Posted by Richard on 2005-08-27 22:14:22