(Edited by Dionysius Lardner, 1855)

Proposition XXXII. Theorem.

[Euclid, ed. Lardner, 1855, on `Google Books`]

(124) | If any side (A B) of a triangle (A B C) be produced, the external angle (F B C) is equal to the sum of the two internal and opposite angles (A and C); and the three internal angles of every triangle taken together are equal to two right angles. |

Through B draw B E parallel to A C (XXXI.) The angle F B E is equal to the internal angle A (XXIX), and the angle E B C is equal to the alternate C (XXIX); therefore the whole external angle F B C is equal to the two internal angles A and C.

The angle A B C with F B C is equal to two right angles (XIII); but F B C is equal to the two angles A and C (first part); therefore the angle A B C together with the angles A and C is equal to two right angles. See Appendix, II.

Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)

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This proposition in other editions:

(125) Cor. 1.—If one angle of a triangle be right, the sum of the other two is equal to a right angle.

(126) Cor. 2.—If one angle of a triangle be equal to the sum of the other two angles, that angle is a right angle.

(127) Cor. 3.—An obtuse angle of a triangle is greater and an acute angle less than the sum of the other two angles.

(128) Cor. 4.—If one angle of a triangle be greater than the sum of the other two it must be obtuse; and if it be less than the sum of the other two it must be acute.

(129) Cor. 5.—If two triangles have two angles in the one respectively equal to two angles in the other, the remaining angles must be also equal.

(130) Cor. 6.—Isosceles triangles having equal vertical angles must also have equal base angles.

(131) Cor. 7.—Each base angle of an isosceles triangle is equal to half the external vertical angle.

(132) Cor. 8.—The line which bisects the external vertical angle of an isosceles triangle is parallel to the base, and

vice versa.(133) Cor. 9.—In a right-angled isosceles triangle each base angle is equal to half a right angle.

(134) Cor. 10.—All the internal angles of any rectilinear figure A B C D E, together with four right angles, are equal to twice as many right angles as the figure has sides.

Take any point F within the figure, and draw the right lines F A, F B, F C, F D, and F E. There are formed as many triangle as the figure has sides, and therefore all their angles taken together are equal to twice as many right angles as the figure has sides (XXXII); but the angles at the point F are equal to four right angles (83); and therefore the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

This is the first corollary in the Elements, and the following is the second.

(135) Cor. 11.—The external angles of any rectilinear figure are together equal to four right angles: for each external angle, with the internal adjacent to it, is equal to two right angles (XIII); therefore all the external angles with all the internal are equal to twice as many right angles as the figure has sides; but the internal angles, together with four right angles, are equal to twice as many right angles as the figure has sides (134). Take from both, the internal angles and the internal remain equal to four right angles.

^{★}★^{★}This corollary is only true of what are calledconvex figures; that is, of figures in which every internal angle is less than two right angles. Some figures, however, have angles which are calledreentrantangles, and which are greater than two right angles. Thus in this figure the angle A B C exceeds two right angles, by the figure K B A, formed by the side B A with the production of the side B C. This angle K B A, is that which in ordinary cases is the external angle, but which in the present instance constitutes a part of the internal angle, and in this case there is no external angle. The angle which is considered as the reentrant angle, and one of the internal angles of the figure is marked with the dotted curve in the figure. See (14).^{★}★^{★}(136) A figure which has no reentrant angle is called aconvex figure.It should be observed, that the first corollary applies to all rectilinear figures, whether convex or not, but the second only to convex figures.

^{★}★^{★}(137) If a figure be not convex each reentrant angle exceeds two right angles by a certain excess, and has no adjacent external angle, while each ordinary angle, together with its adjacent external angle, is equal to two right angles. Hence it follows, that the sum of all the angles internal and external, including the reentrant angles, is equal to twice as many right angles as the figure has sides, together with the excess of every reentrant angle above two right angles. But (134) the sum of the internal angles alone is equal to twice as many right angles as the figure has sides, deducting four; hence the sum of the external angles must be equal to those four right angles, together with the excess of every reentrant angle above two right angles.The sum of the external angles of every convex figure must be the same; and, however numerous the sides and angles be, this sum can never exceed four right angles.

If every pair of alternate sides of a convex figure be produced to meet, the sum of the angles so formed will be equal to 2

n- 8 right angles. This may be proved by showing that each of these angles with two of the external angles is equal to two right angles.^{★}★^{★}(138) Cor. 12.—The sum of the internal angles of a figure is equal to a number of right angles expressed by twice the number of sides, deducing four; also as each reentrant angle must be greater than two right angles, the sum of the reentrant angles must be greater than twice as many right angles as there are reentrant angles. Hence it follows, that twice the number of sides deducting four, must be greater than twice the number of reentrant angles, and therefore that the number of sides deducting two, must be greater than the number of reentrant angles; from which it appears, that the number of reentrant angles in a figure must always be at least three less than the number of sides. There must be therefore at least three angles in every figure, which are each less than two right angles.^{★}★^{★}(139) Cor. 13.—A triangle cannot therefore have any reentrant angle, which also follows immediately from considering that the three angles are together equal to two right angles, while a single reentrant angle would be greater than two right angles.^{★}★^{★}(140) Cor. 14.—No equiangular figure can have a reentrant angle, for if one angle were reentrant all should be so, which cannot be (138).^{★}★^{★}(141) Cor. 15.—If the number of sides in an equiangular figure be given, the magnitude of its angles can be determined. Since it can have no reentrant angle, the sum of its external angles is equal to four right angles; the magnitude of each external angle is therefore determined by dividing four right angles by the number of sides. This being deducted from two right angles, the remainder will be the magnitude of each angle. Thus the fraction whose numerator is 4, and those denominator is the number of sides, expresses the part of a right angle which is equal to the external angle of the figure, and if this fraction be deducted from the number 2, the remainder wil express the internal angle in parts of a right angle. In the notation of arithmetic, ifnbe the number of sides, the external angle is the $\frac{4}{n}$^{th}and the internal angle the $(2-\frac{4}{n})$^{th}of a right angle.^{★}★^{★}(142) Cor. 16.—The sum of the angles of every figure is equal to an even number of right angles. For twice the number of sides is necessarily even, and the even number four being subducted leaves an even remainder. Hence it appears, that no figure can be constructed the sum of whose angles is equal to 3, 5, or 7 right angles, &c.^{★}★^{★}(143) Cor. 17.—If the number of right angles to which the sum of the angles of any figure is equal be given, the number of sides may be found. For since the number of right angles increased by four is equal to twice the number of sides, it follows, that half the number of right angles increased by two is equal to the number of sides.^{★}★^{★}(144) Cor. 18.—If all the angles of a figure be right, it must be a quadrilateral, and therefore a right angled parallelogram. For (141) the magnitude of each external angle is determined in parts of a right angle by dividing 4 by the number of sides; in the present case each external angle must be a right angle, and therefore 4 divided by the number of sides must be 1, and therefore the number of sides must be four. Each of the four angles being right, every adjacent pair is equal to two right angles, and therefore the opposite sides of the figure are parallel.^{★}★^{★}(145) Cor. 19.—The angle of an equilateral triangle is equal to one third of two right angles, or two thirds of a right angle.That one third of two right angles is equal to two thirds of one right angle, easily appears from considering that as three thirds of a right angle is equal to one right angle, six thirds will be equal to two right angles, and one third of this is two thirds of one right angle.

^{★}★^{★}(146) Cor. 20.—To trisect a right angle. Construct any equilateral triangle and draw a line (XXIII), cutting off from the given angle an angle equal to an angle of the equilateral triangle. This angle being two thirds of the whole, if it be bisected, the whole right angle will be trisected.By the combination of bisection and trisection a right angle may be divided into 2, 3, 4, 6, 8, &c equal parts.

N.B. The general problem to trisect

any angleis one which has never been solved by plane Geometry.^{★}★^{★}(147) Cor. 21.—The multisection of a right angle may be extended by means of the angles of the regular polygons.In a regular pentagon the external angle is four fifths of a right angle; the complement of this angle being the fifth of a right angle solves the problem to divide a right angle into five equal parts.

In a regular heptagon the external angle is four sevenths of a right angle, which being divided into four equal parts (IX) gives the seventh of a right angle, and solves the problem to divide a right angle into seven equal parts.

Thus in general the problem of the multisection of a right angle is resolved to that of the construction of the regular polygons, and

vice versa. On this subject the student is referred to the fourth book of the Elements.^{★}★^{★}(148) Cor. 22.—The vertical angle A of a triangle is right, acute or obtuse, according as the line A D which bisects the base B C is equal to, greater or less than half the base B D.1. If the line A D be equal to half the base B D, the triangles A D B and A D C will be isosceles, therefore the angles B A D and C A D will be respectively equal to the angles B and C. The angle A is therefore equal to the sum of B and C, and is therefore (126) a right angle.

2. If A D be greater than B D or D C, the angles B A D and C A D are respectively less than the angles B and C, and therefore the angle A is less than the sum of B and C, and is therefore (128) acute.

3. If A D be less than B D or D C, the angles B A D and C A D are respectively greater than B and C, and therefore the angle A is greater than the sum of B and C, and is therefore (128) obtuse.

^{★}★^{★}(149) Cor. 23.—The line drawn from the vertex A of a triangle bisecting the base B C is equal to, greater or less than half the base, according as the angle A is right, acute, or obtuse.1. Let the angle A be right. Draw A D so that the angle B A D shall be equal to the angle B. The line A D will then bisect B C, and be equal to half of it.

For the angles B and C are together equal to the angle A (125), and since B is equal to B A D, C must be equal to C A D. Hence it follows, (VI) that B D A and C D A are isosceles triangles, and that B D and C D are equal to A D and to each other.

2. Let A be acute, and draw A D bisecting B C. The line A D must be greater than B D or D C; for if it were equal to them the angle A would be right, and if it were less it would be obtuse (148).

3. Let A be obtuse, and draw A D bisecting B C. The line A D must be less than each of the parts B D, D C; for if it were equal to them the angle A would be right, and if it were greater the angle A would be acute (148).

^{★}★^{★}(150) Cor. 24.—To draw a perpendicular to a given right line through its extremity without producing it.Take a part A B from the extremity A, and construct on it an equilateral triangle A C B. Produce B C so that C D shall be equal to A C, and draw D A. This will be the perpendicular required. For since A C bisects B D, and is equal to half of it, the angle D A B is right (148).