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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