Optics

it is impossible and contrary to [Proposition] 32 of the first book; which is also evident in convex surfaces. For since, by definition, every perpendicular line that meets a convex surface is perpendicular to the plane surface tangent to that convex surface at the point of incidence of that line, it is clear that in every convex surface the same impossibility occurs. For if there were a spherical convex surface in which there were an arc $fg$, let it be that a plane surface touches it at point $f$, in which line $hfk$ is drawn, and at point $g$ a plane surface in which there is line $lgm$. It is clear from the preceding that because angles $efk$ and $egm$ are right, having also drawn the chord $fg$, it is clear that angles $efg$ and $egf$ are greater than right angles, which is impossible. Therefore, it is not possible for more than one perpendicular to be drawn from one given point to a plane or convex surface; thus the proposition is clear, since in any other convex surfaces it is to be demonstrated in the same manner.

XXI.

Let $bcd$ be a plane surface, and $a$ a point marked outside it, from which let many lines be drawn to the given surface as it happens, namely $ae, af, ag, ah$, but let $ae$ alone be the perpendicular. I say that line $ae$ is the shortest of all the others. Let lines $ef, eg, eh$ be drawn, and right-angled triangles be constructed; it is thus clear, since by [Proposition] 32 of the first book the right angle is the greatest in any right-angled triangle, that line $ae$, by [Proposition] 19 of the first book, is shorter than any of the lines $af, ag, ah$, and also any others so produced. The proposition is therefore clear in planes. But the same is also clear in convex [surfaces], because if $ae$ is the perpendicular to the convex surface, and $bcdi$ is a plane surface tangent to the convex surface at point $e$, and lines $af, ag, ah$ are drawn to the plane surface, those will all be greater than the perpendicular; but those same [lines] produced to the convex surface are [even] greater. Therefore the proposition is clear.

A geometric diagram labeled with letters A through I. It depicts a point A above a horizontal line segment representing a plane. Several lines descend from A to the plane: AE (a vertical perpendicular line), AF, AG, and AH. The entire figure is contained within a rectangular frame bounded by points B, C, D, and I.

XXII.

Let there be a point given in the air which is $a$, from which let line $ab$ be erected by [Proposition] 12 of the eleventh book upon the underlying plane surface $bcd$, meeting the given surface at point $b$. In surface $bcd$, let line $dc$ be drawn as one pleases, and from point $b$ let a perpendicular be drawn to line $dc$. For let any point $c$ be taken on line $dc$, and let lines $ac$ and $bc$ be drawn. Because, therefore, line $ab$ is erected upon the surface $bcd$, it is clear by the definition of an erected line that angle $abc$ is a right angle; therefore, by the penultimate [proposition] of the first book, the square of line $ac$ is equal to the two squares of lines $ab$ and $bc$. But also the square of line $bc$ is equal to the two squares of $cd$ and $bd$ by the same penultimate [proposition of the first book], because line $bd$ is perpendicular to line $cd$ by hypothesis. Thus the square of line $ac$ is equal to the three squares of the three lines $ab$, $bd$, and $cd$. But the square of line $ad$ is equal to the two squares of the two lines $ab$ and $bd$; the square of line $ac$ is therefore equal to the two squares of the two lines $ad$ and $dc$. Therefore, by the last [proposition] of the first book, angle $adc$ is a right angle. It is clear, therefore, that line $ad$ is perpendicular to line $dc$, which is the proposition.

A geometric diagram illustrating the three-perpendiculars theorem. It shows a vertical line AB meeting a plane at B. On the plane, line BD is drawn perpendicular to line CD. Lines AD and AC are drawn from point A to the plane, creating a 3D geometric construction. Right-angle symbols are marked at points B and D.