Optics

XXIII.

When a single straight line falls upon two parallel plane surfaces, if it is perpendicular to one of them, it will also be perpendicular to the other.

A geometric diagram showing two parallel horizontal lines (surfaces) labeled A-C and B-D, connected by a vertical line A-B. Points are labeled A, B, C, and D.

Let a line $ab$ fall upon two plane and parallel surfaces, meeting one at point $a$ and the other at point $b$. I say that if line $ab$ is perpendicular to one of these surfaces, it will also be perpendicular to the other. Let a straight line $ac$ be drawn from point $a$ in one of those surfaces, and from point $b$ let a line $bd$ be drawn in the other. It is then clear that lines $ac$ and $bd$ are parallel, for if extended infinitely they will not meet, because the surfaces in which they lie do not meet. If, therefore, one of the angles, either $bac$ or $abd$, is a right angle, it is always clear by [Proposition] 29 of the first [book of Euclid] that the other of them will also be a right angle; and since this can be declared in the same way for all lines drawn in the surfaces from points $a$ and $b$, it is evident that line $ab$ makes right angles with each of the lines meeting it produced in either of those surfaces. If, therefore, line $ab$ is perpendicular to one of the surfaces, it is clear that it is perpendicular to the other of them; and this is what was proposed.

XXIIII.

If two surfaces are parallel to one and the same surface, they will be parallel to each other; also, a surface intersecting one of the parallel surfaces will also intersect the other.

A geometric diagram showing three parallel horizontal lines representing surfaces. The top is labeled B-C, the middle A-E, and the bottom H-G. A vertical line labeled I-L-M intersects them. Points are labeled B, C, I, A, E, L, D, K, H, G, M.

Let there be two surfaces $abc$ and $ghk$ parallel to one surface $def$. I say that these two surfaces $abc$ and $ghk$ will necessarily be parallel to each other. For let a line $lm$ be drawn from point $l$ of surface $abc$ perpendicular to that surface by [Proposition] 12 of the eleventh [book], and let it be $lm$. It is then clear by the preceding [proposition] that this line $lm$, extended beyond either of its ends, will itself be perpendicular to surface $ghk$ (which is parallel to surface $abc$) by that same preceding proposition. Therefore, since one line $lm$ stands orthogonally upon the two surfaces $abc$ and $ghk$, it is clear by [Proposition] 14 of the eleventh [book] that those two surfaces, even if extended infinitely, will never meet; they are therefore parallel. Thus the first part of the proposition is clear, and by this and by [Proposition] 2 of this [book], the second part of the proposition is also clear.

XXV.

All perpendicular lines drawn between parallel lines or surfaces are parallel and equal; and if straight lines fall upon parallel lines or surfaces at equal angles, they are equal.

A geometric diagram showing two parallel horizontal lines A-B and F-D. Two vertical lines E-F and G-H connect them. A triangle is formed above with vertex K, and lines K-E and K-G intersecting the top horizontal line. Points are labeled K, A, E, e, g, G, B, F, H, D.

Let there be two parallel lines $ab$ and $cd$, between which let perpendicular lines $ef$ and $gh$ be drawn. I say that lines $ef$ and $gh$ are parallel and equal. For that they are parallel is clear by [Proposition] 28 of the first [book]; that they are also equal is clear by [Proposition] 34 of the first [book]. And it is to be demonstrated in the same way if lines $ab$ and $cd$ are marked in parallel surfaces. But if lines $ef$ and $gh$ fall not perpendicularly, but at equal angles—the lines or surfaces being drawn so that angle $ghc$ is equal to angle $efd$—lines $gh$ and $ef$ will also be equal. For they will meet by [Proposition] 14 of this [book]; let the point of meeting, then, be $k$. Since, therefore, angle $kfh$ is equal to angle $khf$ by hypothesis, the side $kf$ of triangle $kfh$ will be equal to the side $kh$ by [Proposition] 6 of the first [book]. But by [Proposition] 29 and by [Proposition] 16 of the first [book], the side $ke$ of triangle $keg$ will be equal to the side $kg$; therefore there remains line $ef$ equal to line $gh$, which is what was proposed. In the case of lines $ab$ and $cd$ marked in parallel surfaces, the demonstration is the same; thus what was proposed is clear.