Optics

...meet at a point when joined together, are shorter than all lines produced from the same endpoints to another single point upon the same line, which, when joined, contain unequal angles with that same line.

A geometric diagram showing a circular arc marked ABC. Above the arc, points G and D are connected by lines GA and DA to a point A on the arc, where a horizontal tangent line EF passes through. Another point B on the arc is connected to G and D by lines GB and DB. Line GB intersects the tangent line EF at point Z, which is then connected to point D.

Let there be a curved line $abc$, upon the convex side of which, from points $g$ and $d$, lines $da$ and $ga$ fall containing equal angles, such that angle $cag$ is equal to angle $bad$. I say that if other lines are drawn from points $g$ and $d$ to line $abc$, such as $gb$ and $db$, containing unequal angles with line $abc$, then both lines $ga$ and $da$ joined together will be shorter than the two lines $gb$ and $db$ joined together. For let line $ef$ be drawn tangent to the arc $abc$ at point $a$ by [Euclid] III, 16; therefore the angles of contingency, which are $eac$ and $fab$, are equal by III, 15. But angles $gac$ and $dab$ are equal by hypothesis; therefore angles $gae$ and $daf$ will be equal. And to the point where line $gb$ intersects line $ef$, which is $z$, let line $dz$ be drawn; therefore, by the preceding proposition, both lines $ga$ and $da$ are shorter than both lines $gz$ and $dz$, since angle $gza$ is less than angle $gae$, and angle $dzf$ is greater than angle $daf$ by I, 16. But line $gb$ is greater than line $gz$, because the whole is greater than its part, and line $db$ is greater than line $dz$ by I, 19, since angle $dzb$ is greater than the angle of the situated triangle. Therefore the proposition is evident regarding the convex arc of a circle, and must be demonstrated in the same manner in any other cylindrical or pyramidal section according to its convexity; therefore the proposition is clear.

XIX.

Since one straight line exists in two plane surfaces, it is necessary that those two surfaces intersect each other along that line.

A geometric diagram depicting two intersecting rectangular planes. One plane is labeled ABCD and the other is CDEF. They share a common line segment CD as their line of intersection.

Let there be two plane surfaces $abcd$ and $cdef$, in each of which is line $cd$. I say that those two surfaces intersect each other along line $cd$. For if those two surfaces were joined at line $cd$ as at a common boundary in the manner of a single continuous surface, then it is evident that they are parts of one surface and not two surfaces, which is contrary to the hypothesis. But if the surfaces themselves pass through the given line $cd$, and are not joined to it as to a common boundary, it is clear by [Euclid XI], 3—except when they intersect each other—that some line is common to them. Therefore, they either intersect along line $cd$, and the proposition is held, or along another line which contains the given one; and then since that line is common to both the proposed surfaces by the aforementioned third proposition, unless line $cd$ is also common to them by hypothesis, it follows that two plane surfaces, lying between those two lines, would enclose a body, which is impossible and contrary to the supposition; therefore the proposition is clear.

XX.

From one given point in the air, only one perpendicular can be drawn to any underlying plane or convex surface.

A geometric diagram showing a point E in the air above a base rectangular plane with vertices AA, B, CC, and D. Two lines EF and EG descend from point E to the plane. Additional points labeled F, G, H, and KL help define the geometric relationship of the lines to the surface.

Let a plane surface $abcd$ be given, and a point $e$ given in the air. I say that from point $e$ to the underlying surface it is possible for only one perpendicular to be drawn. For if it were impossible [for this to be the case], let two perpendiculars, which are $ef$ and $eg$, be drawn from point $e$ to the given plane surface $abcd$. Because, therefore, lines $ef$ and $eg$ are joined at an angle at point $e$, it is clear by XI, 2, that those two lines are in the same surface; and since those lines are perpendicular to surface $abcd$, the surface in which those lines lie will be erect upon surface $abcd$. Therefore the common section of this surface and surface $abcd$ is line $fg$ by the preceding proposition; thus in triangle $efg$ there are two right angles, namely $efg$ and $egf$, by the definition of a line erect upon a surface; but this is im- [possible].