By Alhazen (lbn al-Haytham), 965-1039 ; A. Mark Smith (editor, translation, commentary)

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Extra info for Alhacen on the principles of reflection. A Critical Edition, with English Translation and Commentary, of Books 4 and 5 of Alhacen’s De aspectibus. Volume One - Introduction and Latin Text ; Volume two - English Translation

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426-427). Line B’G’ will form angle B’G’D’ with line D’T’. At the center of the circle form angle BGE equal to B’G’D’. Point D, where leg GE of that angle intersects the circle, will be the sought-after point of reflection. Alhacen’s proof that D is in fact the point of reflection is based on figure 17b, p. 535. The gist of the proof—and here I will take some liberties with his actual procedure—is as follows. For a start, it is clear by construction that triangles BGD and B’G’D’ are similar. Draw line BT to form angle BTD equal to half of angle AGB.

But we know from proposition 43 that reflection cannot occur from an angle, such as BTA, that is equal to LGH, so T cannot be a legitimate point of reflection. We also know that all angles, such as BDA, that intersect on arc KL outside tangent circle BGA will be more acute than BTA and thus more acute than LGH. Therefore, angle BDA must be smaller than LGH. Likewise, if circle AGB in figure 16b, p. 532, intersects arc KL at bisection-point X, no reflection can occur within the segment of arc KL between X and point C of intersection to the left of it.

Through point Q’ pass a line perpendicular to M’D’, and from point D’ draw line D’T’ to that perpendicular so as to form angle B’T’D’ = angle BGD = onehalf angle AGB. , the radius of the great circle on the mirror). When angle BGD is formed at the l ALHACEN’S DE ASPECTIBUS center of the mirror equal to the resulting angle B’G’D’, point D where leg GD of that angle intersects the circle will be the point of reflection. Now, using this simple limiting case as a guide, let us return to the original case in which AG and BG are unequal, and let us compare this case to the one just discussed.

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