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.
Looks like Bractal's answer works (with Charlie's correction). I had come up with a different answer:
If BD is parallel to XY then draw a parallel to XY through C. The point this parallel intersects with AX is point E. The intersection of EY and DX is point F. BCFE is the desired tile.
If BD is not parallel to XY then call the point where it intersects XY, "O". The intersection of CO and AX is point E. Find point F as above, and again BCFE is the desired tile.
I like Bractal's answer better though--it's far fewer steps. Constructing a parallel to a line through a point takes several steps by itself.
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