Coroll. IV. That the three angles of a strait-lined plain triangle are equal to two right angles; and any side being produced, the external angle will be equal to the two opposite internal angles. For if there be drawn by the vertex of the plain triangle A B C ([fig. 5]) a parallel to any of the sides, as to A B, the angles A and B will be equal to their alternate angles E and F, and the angle C is common. But, by the [10th article], the three angles E, C and F, are equal to two right angles; and therefore the three angles of the triangle are equal to the same; which is the first. Again, the two angles B and D are equal to two right angles, by the 10th article. Wherefore taking away B, there will remain the angles A and C, equal to the angle D; which is the second.
Coroll. V. If the angles A and B be equal, the sides A C and C B will also be equal, because A B and E F are parallel; and, on the contrary, if the sides A C and C B be equal, the angles A and B will also be equal. For if they be not equal, let the angles B and G be equal. Wherefore, seeing G B and E F are parallels, and the angles G and B equal, the sides G C and C B will also be equal; and because C B and A C are equal by supposition, C G and C A will also be equal; which cannot be, by the [11th article].
Coroll. VI. From hence it is manifest, that if two radii of a circle be connected by a strait line, the angles they make with that connecting line will be equal to one another; and if there be added that segment of the circle, which is subtended by the same line which connects the radii, then the angles, which those radii make with the circumference, will also be equal to one another. For a strait line, which subtends any arch, makes equal angles with the same; because, if the arch and the subtense be divided in the middle, the two halves of the segment will be congruous to one another, by reason of the uniformity both of the circumference of the circle, and of the strait line.
The circumferences of circles are to one another as their diameters are.
13. Perimeters of circles are to one another, as their semidiameters are. For let there be any two circles, as, in the first figure, B C D the greater, and E F G the lesser, having their common centre at A; and let their semidiameters be A C and A E. I say, A C has the same proportion to A E, which the perimeter B C D has to the perimeter E F G. For the magnitude of the semidiameters A C and A E is determined by the distance of the points C and E from the centre A; and the same distances are acquired by the uniform motion of a point from A to C, in such manner, that in equal times the distances acquired be equal. But the perimeters B C D and E F G are also determined by the same distances of the points C and E from the centre A; and therefore the perimeters B C D and E F G, as well as the semidiameters A C and A E, have their magnitudes determined by the same cause, which cause makes, in equal times, equal spaces. Wherefore, by the 13th chapter and [6th article], the perimeters of circles and their semidiameters are proportionals; which was to be proved.
In triangles strait lines parallel to the bases are to one another, as the parts of the sides which they cut off from the vertex.
14. If two strait lines, which constitute an angle, be cut by strait-lined parallels, the intercepted parallels will be to one another, as the parts which they cut off from the vertex. Let the strait lines A B and A C, in the [6th figure], make an angle at A, and be cut by the two strait-lined parallels B C and D E, so that the parts cut off from the vertex in either of those lines, as in A B, may be A B and A D. I say, the parallels B C and D E are to one another, as the parts A B and A D. For let A B be divided into any number of equal parts, as into A F, F D, D B; and by the points F and D, let F G and D E be drawn parallel to the base B C, and cut A C in G and E; and again, by the points G and E, let other strait lines be drawn parallel to A B, and cut B C in H and I. If now the point A be understood to be moved uniformly over A B, and in the same time B be moved to C, and all the points F, D, and B be moved uniformly and with equal swiftness over F G, D E, and B C; then shall B pass over B H, equal to F G, in the same time that A passes over A F; and A F and F G will be to one another, as their velocities are; and when A is in F, D will be in K; when A is in D, D will be in E; and in what manner the point A passes by the points F, D, and B, in the same manner the point B will pass by the points H, I, and C; and the strait lines F G, D K, K E, B H, H I, and I C, are equal, by reason of their parallelism; and therefore, as the velocity in A B is to the velocity in B C, so is A D to D E; but as the velocity in A B is to the velocity in B C, so is A B to B C; that is to say, all the parallels will be severally to all the parts cut off from the vertex, as A F is to F G. Wherefore, A F. G F :: A D. D E :: A B. B C are proportionals.
The subtenses of equal angles in different circles, as the strait lines B C and F E (in [fig. 1]), are to one another as the arches which they subtend. For (by [art. 8]) the arches of equal angles are to one another as their perimeters are; and (by [art. 13]) the perimeters as their semidiameters; but the subtenses B C and F E are parallel to one another by reason of the equality of the angles which they make with the semidiameters; and therefore the same subtenses, by the last precedent article, will be proportional to the semidiameters, that is, to the perimeters, that is, to the arches which they subtend.
By what fraction of a strait line the circumference of a circle is made.
15. If in a circle any number of equal subtenses be placed immediately after one another, and strait lines be drawn from the extreme point of the first subtense to the extreme points of all the rest, the first subtense being produced will make with the second subtense an external angle double to that, which is made by the same first subtense, and a tangent to the circle touching it in the extreme points thereof; and if a strait line which subtends two of those arches be produced, it will make an external angle with the third subtense, triple to the angle which is made by the tangent with the first subtense; and so continually. For with the radius A B (in [fig. 7]) let a circle be described, and in it let any number of equal subtenses, B C, C D, and D E, be placed; also let B D and B E be drawn; and by producing B C, B D and B E to any distance in G, H and I, let them make angles with the subtenses which succeed one another, namely, the external angles G C D, and H D E. Lastly, let the tangent K B be drawn, making with the first subtense the angle K B C. I say the angle G C D is double to the angle K B C, and the angle H D E triple to the same angle K B C. For if A C be drawn cutting B D in M, and from the point C there be drawn L C perpendicular to the same A C, then C L and M D will be parallel, by reason of the right angles at C and M; and therefore the alterne angles L C D and B D C will be equal: as also the angles B D C and C B D will be equal, because of the equality of the strait lines B C and C D. Wherefore the angle G C D is double to either of the angles C B D or C D B; and therefore also the angle G C D is double to the angle L C D, that is, to the angle K B C. Again, C D is parallel to B E, by reason of the equality of the angles C B E and D E B, and of the strait lines C B and D E; and therefore the angles G C D and G B E are equal; and consequently G B E, as also D E B is double to the angle K B C. But the external angle H D E is equal to the two internal D E B and D B E; and therefore the angle H D E is triple to the angle K B C, &c.; which was to be proved.