Equal and parallel magnitudes at an unequal distance from the eye are not seen proportionally to the distances.

(general diagram)

(diagram 1)Let there be two magnitudes, AB, GD at an unequal distance from the eye E. (diagram 2) I say that it is not the case that they appear as holding : as GD is to AB [i.e., the appearance of GD from E to the appearance of AB from E] so is BE to ED.

(diagram 3) For let rays AE, EG fall on them, and (diagram 4 = general diagram) with center E and distance EZ let the circular-arc of a circle be inscribed, HZQ. (diagram 5) And so since triangle EZG is larger than section EZH, (diagram 6) while triangle EZD is smaller than section EZQ, (diagram 7) therefore triangle EZG to section EZH has a larger ratio than triangle EZD to section EZQ. (diagram 8) And by alternation triangle EZG to triangle EZD has a larger ratio than section EZH to section EZQ, (diagram 9) and by composition triangle EGD to triangle EZD has a larger ratio than section EHQ to section EZQ. (diagram 10)But as EDG is to triangle EZD, so is GD to DZ. (diagram 11) But GD is equal to AB, and as AB is to DZ, BE is to ED. (diagram 12) Therefore BE to ED has a larger ratio than section EHQ to section EZQ. But as the section is to the section, so is angle HEQ to angle ZEQ. Therefore BE to ED has a larger ratio than the angle HEQ to the ZEQ. And GD is seen from angle HEQ, while AB is seen from ZEQ. Therefore, equal magnitudes are not seen proportional to their distances.from ZEQ. Therefore, equal magnitudes are not seen proportional to their distances.

Note: this proof establishes more than the theorem states. For the theorem merely states that BE : DE is not equal to GED : AEB. But the proof shows

if BE > DE, then BE : DE > GED : AEB

This is one of the basic trigonometric theorems.