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
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.
|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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