Optics

For example: let two lines $ab$ and $cd$ be between two parallel lines, from whose endpoints they are produced, which are $ad$ and $cb$; let them intersect at point $e$ so that line $ae$ is equal to line $eb$, and line $ce$ is equal to $ed$ itself. I say that line $ad$ is equal to line $cb$. For since by the 15th [proposition] of the First [Book of Euclid] angle $aed$ is equal to angle $ceb$, it will be, from the hypothesis and by the 4th [proposition] of the First, that line $ad$ is equal to line $cb$, which is what was proposed.

A geometric diagram showing two horizontal parallel lines. Two intersecting diagonal lines connect the endpoints of the parallel lines (A-B and C-D), crossing at a point E. Letters A, B, C, D, and E label the vertices and intersection point.

XVI.

If straight lines are produced through the endpoints of two parallel and unequal lines, it is necessary that they meet on the side of the smaller line.

Let there be two parallel and unequal lines $ab$ and $cd$, and let line $cd$ be smaller than line $ab$; let lines $ac$ and $bd$ be produced through their endpoints. I say that those lines $ac$ and $bd$ will meet beyond line $cd$. For let line $cd$ be produced beyond point $d$ to point $e$, and by the third [proposition] of the First let line $ce$ be made equal to line $ab$, and let line $be$ be drawn. Here, then, since line $be$ by the 33rd [proposition] of the First is parallel to line $ac$, therefore by the 2nd [proposition] of this [book], since line $bd$ meets line $be$ at point $b$, it is evident that it meets line $ac$, which is parallel to line $be$. But it is also necessary that they meet on the side of line $cd$, which is smaller than line $ab$, by the 14th [proposition] of this [book] or by the 2nd [proposition] of the Sixth; therefore the proposal is evident, for the point of meeting, which is $f$, will be beyond line $cd$.

A geometric diagram with a base line A-F. Parallel lines A-C and B-D are shown, but converging towards a future intersection point F. Additional construction lines extend to a point E.

XVII.

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

Let there be a straight line $abcf$, and let there be two points $d$ and $g$, from which two lines $gb$ and $db$ are produced upon line $abcf$, containing equal angles, such that angle $abg$ is equal to angle $cbd$. I say that if lines drawn from points $d$ and $g$ to some other point of line $abcf$, which is $c$, contain unequal angles, such that angle $gca$ is smaller than angle $fcd$, then lines $gb$ and $bd$ joined together [are shorter] than the two lines $gc$ and $dc$ joined together. For let a perpendicular be drawn from point $g$ onto line $af$ by the 12th [proposition] of the First, which is $gh$; and let line $gh$ be produced beyond point $h$, and let $db$ be produced until it meets the produced line $gh$ (they will indeed meet by the 14th [proposition] of this [book]); therefore let the point of meeting be $k$, and let line $kc$ be joined. And since angle $dbc$ is equal to angle $gbh$ by hypothesis, and [equal] to angle $hbk$ by the 15th [proposition] of the First, it is clear that angle $hbk$ is equal to $gbh$. But angles $ghb$ and $khb$ are equal because they are right angles; therefore, by the 32nd [proposition] of the First, the triangles $ghb$ and $khb$ are also equiangular; therefore, by the 4th [proposition] of the Sixth, since line $hb$ is common and equal to itself, line $gb$ will be equal to line $kb$, and line $gh$ equal to line $hk$. And by the same reasoning, by the 4th [proposition] of the First, line $gc$ will be equal to line $kc$. Because indeed by the 20th [proposition] of the First, line $kd$ in triangle $kdc$ is less than both lines $dc$ and $kc$ joined together, and line $gb$ is equal to line $bk$, and line $gc$ is equal to line $kc$, it is clear that both lines $gb$ and $db$ joined together are less than both lines $dc$ and $gc$ joined together. Likewise, it is to be demonstrated concerning any lines produced from points $g$ and $d$ to line $af$; therefore the proposal is evident.

A complex geometric diagram showing a horizontal line labeled A-B-C-FF. Points G and D are above the line. A vertical line drops from G through the horizontal line at H to a point K. Several lines intersect at points B and C on the horizontal axis.

XVIII.

Straight lines containing equal angles with a convex line, which at one