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