In this applet points C and O' are moveable. If the diagonal through B is extended beyond the image X of the far left corner of the checkerboard, it meets the horizon in the point OO. (Any one of the parallel diagonals will pass through OO). Arguing with similar triangles, we can show that the distance from OO to C is equal to the distance from O' to C' and is therefore equal to the distance from the picture of the eye used in the construction.

The argument runs as follows: Equating ratos of corresponding sides in the similar triangles   H' C' O'  and  H' B A  gives

|H' C'| / |B H'| = |O' C'| / |A B|.

In the similar triangles  X C OO  and  X A B  the ratios of the altitudes must be the same as the ratio of the bases. This gives

|H' C'| / |B H'| = |OO C| / |A B|.

It follows that |OO C| = |O'C'|.

This applet was prepared with JavaSketchpad, a World-Wide-Web component of The Geometer's Sketchpad. Copyright ©1990-1998 by Key Curriculum Press, Inc. All rights reserved. Portions of this work were funded by the National Science Foundation (awards DMI 9561674 & 9623018).

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