Prop. 1: From plane mirrors and convex mirrors and concave mirrors the sight-lines are reflected in equal angles.

(diagram 1) Let there be an eye B, plane mirror AG, let there move a sight-line BK from the eye and let it be reflected to D. I say, in fact, the angle E is equal to Z. Let there be drawn altitudes to the mirror BG, DA. And so as BG is to GK, DA is to AK. For this was supposed in the definitions. Therefore, triangle BGK is similar to triangle DAK. Therefore angle E is equal to angle Z. For the triangles are equal angled.
(diagram 2) Let there be a convex mirror AKG, sight-line BK reflected to D. I say that angle E, Q is equal to Z, L. I placed a plane mirror NM along there. Therefore, angle E is equal to Z. But also Q is equal to L. For MN is tangent. Therefore the whole E, Q is equal to the whole LZ.
(diagram 3) Let there be again a concave mirror AKG, sight-line BK reflected to D. I say that angle E angle is equal to Z. For when a plane mirror is placed along there, angle Q, E becomes equal to angle Z, L. But Q is equal to L. Therefore, the remainder E is equal to Z.

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