Euclid, Book I, as translated by Thomas L. Heath

(58)   In the solution of this problem it is assumed that the two circles intersect, inasmuch as the vertex of the equilateral triangle is a point of intersection. This, however, is sufficiently evident if it be considered that a circle is a continued line which includes space, and that in the present instance each circle passing through the centre of the other must have a part of its circumference within that other, and a part without it, and must therefore intersect it.

It follows from the solution, that as many different equilateral triangles can be constructed on the same right line as there are points in which the two circles intersect. It will hereafter be proved that two circles cannot intersect in more than two points, but for the present this may be taken for granted.

Since there are but two points of intersection of the circles, there can be but two equilateral triangles constructed on the same finite right line, and these are placed on opposite sides of it, their vertices being at the points C and F.

After having read the first book of the elements, the student will find no difficulty in proving that the triangles C F E and C D F are equilateral. These lines are not in the diagram, but may easily be supplied.

^★★^★ The different positions which the given right line and given point may have with respect to each other, are apt to occasion such changes in the diagram as to lead the student into error in the execution of the construction for the solution of this problem.

Hence it is necessary that in solving this problem, the student should be guided by certain general directions, which are independent of any particular arrangement which the several lines concerned in the solution may assume. If the student is governed by the following general directions, no change which the diagram can undergo will mislead him.

1° The given point is to be joined with either extremity of the given right line. (Let us call the extremity with which it is connected, the connected extremity of the given right line; and the line so connecting them, the joining line.)

2° The centre of the first circle is the connected extremity of the given right line; and its radius, the given right line.

3° The equilateral triangle may be constructed on either side of the joining line.

4° The side of the equilateral triangle which is produced to meet the circle, is that side which is opposite to the given point, and it is produced through the centre of the first circle till it meets its circumference.

5° The centre of the second circle is that vertex of the triangle which is opposite to the joining line, and its radius is made up of that side of the triangle which is opposite to the given point, and its production which is the radius of the first circle. So that the radius of the second circle is the sum of the side of the triangle and the radius of the first circle.

6° The side of the equilateral triangle which is produced through the given point to meet the second circle, is that side which is opposite to the connected extremity of the given right line, and the production of this side is the line which solves the problem; for the sum of this line and the side of the triangle is the radius of the second circle, but also the sum of the given right line (which is the radius of the first circle) and the side of the triangle is equal to the radius of the second circle. The side of the triangle being taken away the remainders are equal.

As the given point may be joined with either extremity, there may be two different joining lines, and as the triangle may be constructed on either side of each of these, there may be four different triangles; so the right line and the point being given, there are four different constructions by which the problem may be solved.

If the student inquires further, he will perceive that the solution may be effected also by producing the side of the triangle opposite the given point, not through the extremity of the right line but through the vertex of the triangle. The various consequences of this variety in the construction we leave to the student to trace.

(60)   By the second proposition a right line of a given length can be inflected from a given point P upon any given line A B. For from the point P draw a right line of the given length (II), and with P as centre, and that line as radius, describe a circle. A line drawn from P to any point C, where this circle meets the given line A B, will solve the problem.

By this proposition the first may be generalized; for an isosceles triangle may be constructed on a given line as base, and having its side of a given length. The construction will remain unaltered, except that the radius of each of the circles will be equal to the length of the side of the proposed triangle. If this length be not greater than half the base, the two circles will not intersect, and no triangle can be constructed, as will appear hereafter.

By a similar construction, the less might be produced until it equal the greater. From an extremity of the less let a line equal to the greater be drawn, and a circle be described with this line as radius. Let the less be produced to meet this circle.

In the demonstration of this proposition, the converse of the eighth axiom (50) is assumed. The axiom states, that ‘if two magnitudes coincide they must be equal.’ In the proposition it is assumed, that if they be equal they must under certain circumstances coincide. For when the point D is placed on A, and the side D E on A B, it is assumed that the point E must fall on B, because A B and D E are equal. This may, however, be proved by the combination of the eighth and ninth axioms; for if the point E did not fall upon B, but fell either above or below it, we should have either E D equal to a part of B A, or B A equal to a part of E D. In either case the ninth axiom would be contradicted, as we should have the whole equal to its part.

The same principle may be applied in proving that the side D F will fall upon A C, which is assumed in Euclid's proof.

In the superposition of the triangles in this proposition, three things are to be attended to:

1° The vertices of the equal angles are to be placed one on the other.

2° Two equal sides to be placed one on the other.

3° The other two equal sides are to be placed on the same side of those which are laid one upon the other.

From this arrangement the coincidence of the triangles is inferred.

It should be observed, that this superposition is not assumed to be actually effected, for that would require other postulates besides the three already stated; but it is sufficient for the validity of the reasoning, if it be conceived to be possible that the triangles might be so placed. By the same principle of superposition, the following theorem must be easily demonstrated, ‘If two triangles have two angles in one respectively equal to two angles in the other, and the sides lying between those angles also equal, the remaining sides and angles will be equal, and also the triangles themselves will be equal.’ See prop. xxvi.

This being the first theorem in the Elements, it is necessarily deduced exclusively from the axioms, as the first problem must be from the postulates. Subsequent theorems and problems will be deduced from those previously established.

In the construction for this proposition it is necessary that the part of the greater side which is cut off equal to the less, should be measured upon the greater side B A from vertex (B) of the equal angle, for otherwise the fourth proposition could not be applied to prove the equality of the part with the whole.

It may be observed generally, then when a part of one line is cut off equal to another, it should be distinctly specified from which extremity the part is to be cut.

This proposition is what is called by logicians the converse of the fifth. It cannot however be inferred from it by the logical operation called conversion; because, by the established principles of Aristotelian logic, an universal affirmative admits no simple converse. This observation applies generally to those propositions in the Elements which are converses of preceding ones.

The demonstration of the sixth is the first instance of indirect proof which occurs in the Elements. The force of this species of demonstration consists in showing that a principle is true, because some manifest absurdity would follow from supposing it to be false.

This kind of proof is considered inferior to direct demonstration, because it only proves that a thing must be so, but fails in showing why it must be so; whereas direct proof not only shows that the thing is so, but why it is so. Consequently, indirect demonstration is never used, except where no direct proof can be had. It is used generally in proving principles which are nearly self-evident, and in the Elements if oftenest used in establishing the converse propositions. Examples will be seen in the 14th, 19th, 25th and 40th propositions of this book.

This proposition seems to have been introduced into the Elements merely for the purpose of establishing that which follows it. The demonstration is that form of argument which logicians call a dilemma, and a species of argument which seldom occurs in the Elements. If two triangles whose conterminous sides are equal could stand on the same side of the same base, the vertex of the one must necessarily either fall within the other or without it, or on one of the sides of it: accordingly, it is successively proved in the demonstration, that to suppose it in any of these positions would lead to a contradiction in terms. It is not supposed that the vertex of the one could fall on the vertex of the other; for that would be supposing the two triangles to be one and the same, whereas they are, by hypothesis, different.

In the Greek text there is but one (the first) of the cases of this proposition given. It is however conjectured, that the second case must have been formerly in the text, because it is the only instance in which Euclid uses that part of the fifth proposition which proves the equality of the angles below the base. It is argued, that there must have been some reason for introducing into the fifth a principle which follows at once from the thirteenth; and that none can be assigned except the necessity of the principle in the second case of the seventh. The third case required to be mentioned only to preserve the complete logical form of the argument.

Note regarding Hypotheses

[This note occurs as a footnote in printed editions, referenced in the course of the proof of Proposition VII.]

The hypothesis means the supposition; that is, the part of the enunciation of the proposition in which something is supposed to be granted true, and from which the proposed conclusion is to be inferred. Thus in the seventh proposition the hypothesis is, that the triangles stand on the same side of their base, and that their conterminous sides are equal, and the conclusion is a manifest absurdity, which proves that the hypothesis must be false.

In the fourth proposition the hypothesis is, that two sides and the included angle of one triangle are respectively equal to two sides and the included angle of the other; and the conclusion deduced from this hypothesis is, that the remaining side and angles in the one triangle are respectively equal to the remaining side and angles in the other triangle.

(69)   It is evident that in this case all the angles and sides of the triangles are respectively equal each to each, and that the triangles themselves are equal. This appears immediately by the eighth axiom.

In order to remove from the threshold of the Elements a proposition so useless, and, to the younger students, so embarrassing as the seventh, it would be desirable that the eighth should be established independently of it. There are several ways in which this might be effected. The following proof seems liable to no objection, and establishes the eighth by the fifth.

Let the two equal bases be so applied one upon the other that the equal sides shall be conterminous, and that the triangles shall lie at opposite sides of them, and let a right line be conceived to be drawn joining the vertices.

1° Let this line intersect the base.

Let the vertex F fall at G, the side E F in the position A G, and D F in the position C G. Hence B A and A G being equal, the angles G B A and B G A are equal (V). Also C B and C G being equal, the angles C G B and C B G are equal (V). Adding these equals to the former, the angles A B C and A G C are equal; that is, the angles E F D and A B C are equal.

2° Let the line G B fall outside the coincident bases.

The angles G B A and B G A, and also B G C and G B C are proved equal as before; and taking the latter from the former, the remainders, which are the angles A G C and A B C, are equal, but A G C is the angle F.

3° Let the line B G pass through either extremity of the base.

In this case it follows immediately (V) that the angles A B C and A G C are equal; for the lines B C and C G must coincide with B G, since each has two points upon it (52).

Hence in every case the angles B and F are equal.

This proposition is also sometimes demonstrated as follows.

Conceive the triangle E F D to be applied to A B C, as in Euclid's proof. Then because E F is equal to A B, the point F must be in the circumference of a circle described with A as centre, and A B as radius. And for the same reason, F must be on a circumference with the centre C, and the radius C B. The vertex must therefore be at the point where these circles meet. But the vertex B must be also at that point; wherefore &c.

By this proposition an angle may be divided into 4. 8, 16 &c. equal parts, or, in general, into any number of equal parts which is expressed by a power of two.

It is necessary that the equilateral triangle be constructed on a different side of the joining line D E from that on which the given angle is placed, lest the vertex F of the equilateral triangle should happen to coincide with the vertex A of the given angle; in which case there would be no joining line F A, and therefore no solution.

In these cases, however, in which the vertex of the equilateral triangle does not coincide with that of the given angle, the problem can be solved by constructing the equilateral triangle on the same side of the joining line D E with the given angle. Separate demonstrations are necessary for the two positions which the vertices may assume.

1. Let the vertex of the equilateral triangle fall within that of the given angle.

The demonstration already given will apply to this without any modification.

2. Let the vertex of the given angle fall within the equilateral triangle.

The line F A produced will in this case bisect the angle; for the three sides of the triangle D F A are respectively equal to those of the triangle E F A. Hence the angles D F A and E F A are equal (VIII). Also, in the triangles D F G and E F G the sides D F and E F are equal, the side G F is common, and the angles D F G and E F G are equal; hence (IV) the bases D G and E G are equal, and also the angles D G A and E G A. Again, in the triangles D G A and E G A the sides D G and E G are equal, A G is common, and the angles at G are equal; hence (IV) the angles D A G and E A G are equal, and therefore the angle B A C is bisected by A G.

It is evident, that an isosceles triangle constructed on the joining line D E would equally answer the purpose of the solution.

In this and the following proposition an isosceles triangle would answer the purposes of the solution equally with an equilateral. In fact, in the demonstrations the triangle is contemplated merely as isosceles: for nothing is inferred from the equality of the base with the sides.

Cor.—By help of this problem it may be demonstrated, that two straight lines cannot have a common segment.

It it be possible, let the two straight lines A B C, A B D have the segment A B common to both of them. From the point B draw B E at right angles to A B; and because A B C is a straight line, the angle C B E is equal to the angle E B A; in the same manner, because A B D is a straight line, the angle D B E is equal to the angle E B A; wherefore the angle D B E is equal to the angle C B E, the less to the greater, which is impossible; therefore the two straight lines cannot have a common segment.

If the given point be at the extremity of the given right line, it must be produced, in order to draw the perpendicular by this construction.

In a succeeding article, the student will find a method of drawing a perpendicular through the extremity of a line without producing it.

The corollary to this proposition is useless, and is omitted in some editions.

It is equivalent to proving that a right line cannot be produced through its extremity in more than one direction, or that it has but one production.

In this proposition it is necessary that the right line A B be indefinite in length, for otherwise it might happen that the circle described with the centre C and the radius C X might not intersect it in two points, which is essential to the solution of the problem.

It is assumed in the solution of this problem, that the circle will intersect the right line in two points. The centre of the circle being on one side of the given right line, and a part of the circumference (X) on the other, it is not difficult to perceive that a part of the circumference must also be also on the same side of the given line with the centre, and since the circle is a continued line it must cross the right line twice. The properties of the circle form the subject of the third book, and those which are assumed here will be established in that part of the Elements.

The following questions will afford the student useful exercise in the application of the geometrical principles which have been established in the last twelve propositions.

(74)	In an isosceles triangle the right line which bisects the vertical angle also bisects the base, and is perpendicular to the base.

For the two triangles into which it divides the isosceles, there are two sides (those of the isosceles) equal, and a side (the bisector) common, and the angles included by those sides equal, being the parts of the bisected angle; hence (IV) the remaining sides and angles are respectively equal; that is, the parts into which the base is divided by the bisector are equal, and the angle which the bisector makes with the base are equal. Therefore it bisects the base, and is perpendicular to it.

It is clear that the isosceles triangle itself is bisected by the bisector of its vertical angle, since the two triangles are equal.

(75)	It follows also, that in an isosceles triangle the line which is drawn from the vertex to the middle point of the base bisects the vertical angle, and is perpendicular to the line.

For in this case the triangle is divided into two triangles, which have their three sides respectively equal each to each, and the property is established by (VIII)

(76)	If in a triangle the perpendicular from the vertex on the base bisect the base, the triangle is isosceles.

For in this case in the two triangles into which the whole is divided by the perpendicular, there are two sides (the parts of the base) equal, one side (the perpendicular) common, and the included angles equal, being right. Hence (IV) the sides of the triangle are equal.

(77)	To find a point which is equidistance from the three vertical points of a triangle A B C.

Bisect the sides A B and B C at D and E (X), through the points D and E draw perpendiculars, and produce them until they meet at F. The point F is at equal distances from A, B and C.

For draw F A, F B, F C. B F A is isosceles by (76), and for the same reason B F C is isosceles. Hence it is evident that F A, F C, and F B are equal.

(78)	Cor.—Hence F is the centre, and F A the radius of a circle circumscribed about the triangle.

(79)	In a quadrilateral formed by two isosceles triangles A C B and A D B constructed on different sides of the same base, the diagonals intersect at right angles, and that which is the common base of the isosceles triangles is bisected by the other.

For in the triangles C A D and C B D the three sides are equal each to each, and therefore (VIII) the angles A C E and B C E are equal. The truth of the proposition therefore follows from (74)

(80)	Hence it follows that the diagonals of a lozenge bisect each other at right angles.

(81)	It follows from (76) that if the diagonals of a quadrilateral bisect each other at right angles it is a lozenge.

The words ‘makes angles with it,’ are introduced to exclude the case in which the line A B is at the extremity of B C.

(83)   From this proposition it appears, that if several right lines stand on the same right line at the same point, and make angles with it, all the angles taken together are equal to two right angles.

Also if two right lines intersecting one another make angles, these angles taken together are equal to four right angles.

The lines which bisect the adjacent angles A B C and A B D are at right angles; for the angle under these lines is evidently half the sum of the angles A B C and A B D.

If several right lines diverge from the same point, the angles into which they divide the surrounding space are together equal to four right angles.

(84)   When two angles as A B C and A B D are togther equal to two right angles, they are said to be supplemental, and one is called the supplement of the other.

(85)   If two angles as C B A and E B A are together equal to a right angle, they are said to be complemental, one one is said to be the complement of the other.

In the enunciation of this proposition, the student should be cautious not to overlook the condition that the two right lines C B and B E forming angles, which are together equal to two right angles, with B A lie at opposite sides of B A. They might form angles together equal to two right angles with B A, yet not lie in the same continued line, if as in this figure they lay at the same side of it. It is assumed in this proposition that the line C B has a production. This is however granted by Postulate 2.

This proof may shortly be expressed by saying, that opposite angles are equal, because they have a common supplement (84).

It is evident that angles which have a common supplement or complement (85) are equal, and that if they be equal, their supplements and complements must also be equal.

(88)   The converse of this proposition may easily be proved, scil. If four lines meet at a point, and the angles vertically opposite be equal, each alternate pair of lines will be in the same right line. For if C E A be equal to B E D, and also C E B to A E D, it follows that C E A and C E B together are equal to B E D and A E D together. But all the four are together equal to four right angles (83), and therefore C E A and C E B are together equal to two right angles, therefore (XIV) A E and A B are in one continued line. In like manner it may be proved, that C E and D E are in one line.

(90)   Cor. 1.—Hence it follows, that each angle of a triangle is less than the supplement of either of the other angles (84). For the external angle is the supplement of the adjacent internal angle (XIII).

(91)   Cor. 2.—If one angle of a triangle be right or obtuse, the others must be acute. For the supplement of a right or obtuse angle is right or acute (82), and each of the other angles must be less than this supplement, and must therefore be acute.

(92)   Cor. 3.—More than one perpendicular cannot be drawn from the same point to the same right line. For if two lines be supposed to be drawn, one of which is perpendicular, they will form a triangle having one right angle. The other angles must therefore be acute (91), and therefore the other line is not perpendicular.

(93)   Cor. 4.—If from any point a right line be drawn to a given right line, making with it an acute and obtuse angle, and from the same point a perpendicular be drawn, the perpendicular must fall at the side of the acute angle. For otherwise a triangle would be formed having a right and an obtuse angle, which cannot be (91).

(94)   Cor. 5.—The equal angles of an isosceles triangle must be both acute.

This proposition and the sixteenth are included in the thirty-second. which proves that the three angles are together equal to two right angles.

This proposition might also be proved by producing the lesser side A B, and taking A E equal to the greater side. In this case the angle A E C is equal to A C E (V), and therefore greater than A C B. But A B C is greater than A E C (XVI), and therefore A B C is greater than A C B

This proposition holds the same relation to the sixth, as the preceding does to the fifth. The four might be thus combined: one angle of a triangle is greater or less than another, or equal to it, according as the side opposed to the one is greater or less than, or equal to the side opposed to the other, and vice versa.

The student generally feels it difficult to remember which of the two, the eighteenth or nineteenth, is proved by construction, and which indirectly. By referring them to the fifth and sixth the difficulty will be removed.

This proposition is sometimes proved by bisecting the angle A. Let A E bisect it. The angle B E A is greater than E A C, and the angle C E A is greater than E A B (XVI); and since the parts of the angle A are equal, it follows, that each of the angles E is greater than each of the parts of A; and thence, by (XIX), it follows that B A is greater than B E, and A C greater than C E, and therefore that the sum of the former is greater than the sum of the latter.

The proposition might likewise be proved by drawing a perpendicular from the angle A on the side B C; but these methods seem inferior in clearness and brevity to that of Euclid.

Some geometers, among whom may be reckoned Archimedes, ridicule this proposition as being self evident, and contend that it should therefore be one of the axioms. That a truth is considered self evident is, however, not a sufficient reason why it should be adopted as a geometrical axiom (57).

(99)   It follows immediately from this proposition, that the difference of any two sides of a triangle is less than the remaining side. For the sides A C and B C taken together are greater than A B; let the side A C be taken from both, and we shall have the side B C greater than the remainder upon taking A C from A B; that is, then the difference between A B and A C.

In this proof we assume something more than is expressed in the fifth axiom. For we take for granted, that if one quantity (a) be greater than another (b), and that equals be taken from both, the remainder of the former (a) will be grater than the remainder of the latter (b). This is a principle which is frequently used, though not directly expressed in the axiom (55).

^★★^★ By the thirty-second proposition it will follow, that the angle B D C exceeds the angle A by the sum of the angles A B D and A C D. For the angle B D C is equal to the sum of D E C and D C E; and, again, the angle D E C is equal to the sum of the angles A and A B E. Therefore the angle B D C is equal to the sum of A, and the angles A B D and A C D.

^★★^★ In this solution Euclid assumes that the two circles will have at least one point of intersection. To prove this, it is only necessary to show that a part of one of the circles will be within, and another part without the other (58).

Since D E and E K or E L are together greater than D K, it follows, that D L is greater than the radius of the circle K G, and therefore the point L is outside the circle. Also, since D K and E K are together greater than D E, if the equals E K and E H be taken from both, D H is less than D K, that is, D H is less than the radius of the circle, and therefore the point H is within it. Since the point H is within the circle and L without it, the one circle must intersect the other.

It is evident, that if the sum of the lines B and C were equal to the line A, the points H and K would coincide; for then the sum of D K and K E would equal D E. Also, if the sum of A and C were equal to B, the points K and L would coincide; for then D K would be equal to E K and D E, or to L D. It will hereafter appear, that in the former case the circles would touch externally, and in the latter internally.

If the line A were greater than the sum of B and C, it is easy to perceive that the circle would not meet, one being wholly outside the other; and if B were greater than the sum of A and C, they would not meet, one being wholly within the other.

If the three right lines A B C be equal, this proposition becomes equivalent to the first, and the solution will be found to agree exactly with that of the first.

It is evident that the eleventh proposition is a particular case of this

In this demonstration it is assumed by Euclid, that the points A and G will be on different sides of B C, or, in other words, that A H is less than A G or A C. This may be proved thus:—The side A C not being less than A B, the angle A B C cannot be less than the angle A C B (XVIII). But the angle A B C must be less than the angle A H C (XVI); therefore the angle A C B is less than A H C, and therefore A H less than A C or A G (XIX).

In the construction for this proposition Euclid has omitted the words ‘with the side which is not the greater.’ Without these it would not follow that the point G would fall below the base B C, and it would be necessary to give demonstrations for the cases in which the point G falls on, or above the base B C. On the other hand, if these words be inserted, it is necessary in order to give validity to the demonstration, to prove as above, that the point G falls below the base.

If the words ‘with the side not the greater’ be not inserted, the two omitted cases may be proved as follows:

If the point G fall on the base B C, it is evident that B G is less than B C (51).

If G fall above the base B C, let it be at G′. The sum of the lines B G′ and A G′ is less than the sum of A C and C B (XXI). The equals A C and A G′ being taken away, there will remain B G′ less than B C.

This proposition might be proved directly thus: On the greater side B C take B G equal to the lesser side E D, and on B G construct a triangle B H G equilateral with E F D. Join A H and produce H G to I.

The angle H will then be equal to the angle F.

1° Let B G be greater than B K.

Since B A and B H are equal, the angles B A H and B H A are equal (V). Also since H G is equal to A C, it is greater than A I, and therefore H I is greater than A I, and therefore the angle H A I is greater than the angle A H I (XVIII). Hence, if the equal triangles B H A and B A H be added to these, the angle B A C will be found greater than the angle B H G, which is equal to F.

2° If B G be not greater than B K, it is evident that the angle H is less than the angle A.

The twenty-fourth and twenty-fifth propositions are analogous to the fourth and eighth, in the same manner as the eighteenth and nineteenth are to the fifth and sixth . The four might be announced together thus:

If two triangles have two sides of the one respectively equal to two sides of the other, the remaining side of the one will be greater or less than, or equal to the remaining side of the other, according as the angle opposed to it in the one is greater or less than, or equal to the angle opposed to it in the other, or vice versa.

In fact, these principles amount to this, that if two lines of given lengths be placed so that one pair of extremities coincide, and so that in their initial position the lesser line is placed upon the greater, the distance between the extremities will then be the difference of the lines. If they be opened as to form a gradually increasing angle, the line joining their extremities will gradually increase, until the angle they include becomes equal to two right angles, when they will be in one continued line, and the line joining their extremities is their sum. Thus the major and minor limits of this line is the sum and difference of the given lines. This evidently includes the twentieth proposition.

It is evident that the triangles themselves are equal in every respect.

^★★^★ (106)   Cor. 1.—From this proposition and the principles previously established, it easily follows, that a line being drawn from the vertex of a triangle to the base, if any two of the following equalities be given (except the first two), the others may be inferred.

1° The equality of the sides of the triangle.

2° The equality of the angles at the base.

3° The equality of the angles under the line drawn, and the base.

4° The equality of the angles under the line drawn, and the sides.

5° The equality of the segments of the base.

Some of the cases of this investigation have already been proved (74), (75), (76). The others present no difficulty, except in the case where the fourth and fifth equalities are given to infer the others. This case may be proved as follows.

If the line A D which bisects the vertical angle (A) of a triangle also bisect the base B C, the triangle will be isosceles; for produce A D so that D E shall be equal to A D, and join E C. In the triangles D C E and A D B the angles vertically opposed at D are equal, and also the sides which contain them; therefore (IV) the angles B A D and D E C are equal, and also the sides A B and E C. But the angle B A D is equal to D A C (hyp.); and therefore D A C is equal to the angle E, therefore (VI) the sides A C and E C are equal. But A B and E C have already been proved equal, and therefore A B and A C are equal.

^★★^★ (107)   The twenty-sixth proposition furnishes the third criterion which has been established in the Elements for the equality of two triangles. It may be observed, that in a triangle there are six quantities which may enter into consideration, and in which two triangles may agree or differ; viz. the three sides and the three angles. We can in most cases infer the equality of two triangles in every respect, if they agree in any three of those six quantities which are independent of each other. To this, however, there are certain exceptions, as will appear by the following general investigation of the question.

When two triangles agree in three of the six quantities already mentioned, these three must be some of the six following combinations:

1° Two sides and the angle between them.

2° Two angles and the side between them.

3° Two sides, and the angle opposed to one of them.

4° Two angles, and the side opposed to one of them.

5° The three sides.

6° The three angles.

The first case has been established in the fourth, and the second and fourth in the twenty-sixth proposition. The fifth case has been established by the eighth, and in the sixth case the triangles are not necessarily equal. In this case, however, the three data are not independent, for it will appear by the thirty-second proposition, that any one angle of a triangle can be inferred from the other two.

The third is therefore the only case which remains to be investigated.

^★★^★ (108)   3° To determine under what circumstance two triangles having two sides equal each to each, and the angles opposed to one pair of equal sides equal, shall be equal in all respects. Let the sides A B and B C be equal to D E and E F, and the angle A be equal to the angle D. If the two angles B and E be equal, it is evident that the triangles are in every respect equal by (IV), and that C and F are equal. But if B and E be not equal, let one B be greater than the other E; and from B let a line B G be drawn, making the angle A B G equal to the angle E. In the triangles A B G and D E F, the angles A and A B G are equal respectively to D and E, and the side A B is equal to D E, therefore (XXVI) the triangles are in every respect equal; and the side B G is equal to E F, and the angle B G A equal to the angle F. But since E F is equal to B C, B G is equal to B C, and therefore (V) B G C is equal to B C G, and therefore C and B G A or F are supplemental.
(109)   Hence, if two triangles have two sides in the one respectively equal to two sides in the other, and the angles opposed to one pair of equal sides equal, the angles opposed to the other equal sides will be either equal or supplemental.

^★★^★ (110)   Hence it follows, that if two triangles have two sides respectively equal each to each, and the angles opposed to one pair of equal sides equal, the remaining angles will be equal, and therefore the triangles will be in every respect equal, if there be any circumstance from which it may be inferred that the angles opposed to the other pair of equal sides are of the same species.

(Angles are said to be of the same species when they are both acute, both obtuse, or both right).

For in this case, if they be not right they cannot be supplemental, and must therefore be equal (109), in which case the triangles will be in every respect equal, by (XXVI).

If they be both right, the triangles will be equal by (108); because in that case G and C being right angles, B G must coincide with B C, and the triangle B G A with B C A; but the triangle B G A is equal to E F D, therefore &c.

^★★^★ (111)   There are several circumstances which may determine the angles opposed to the other pair of equal sides to be of the same species, and therefore which will determine the equality of the triangles; amongst which are the following:

If one of the two angles opposed to the other pair of equal side be right; for a right angle is its own supplement.

If the angles which are given equal be obtuse or right; for then the other angles must be all acute (91), and therefore of the same species.

If the angles which are included by the equal sides be both right or obtuse; for then the remaining angles must be both acute.

If the equal sides opposed to angles which are not given equal be less than the other sides, these angles must be both acute (XVIII).

In all these cases it may be inferred, that the triangles are in every respect equal.

It will appear by prop. 38, that if two triangles have two sides respectively equal, and the included angles supplemental, their areas are equal.

(The area of a figure is the quantity of surface within its perimeter).

(112)   If several right lines be drawn from a point to a given right line.

1° The shortest is that which is perpendicular to it.

2° Those equally inclined to the perpendicular are equal, and vice versa.

3° Those which meet the right line at equal distances from the perpendicular are equal, and vice versa.

4° Those which make greater angles with the perpendicular are greater, and vice versa.

5° Those which meet the line at greater distances from the perpendicular are greater, and vice versa.

6° More than two equal right lines cannot be drawn from the same point to the same right line.

The student will find no difficulty in establishing these principles.

^★★^★ (113)   If any number of isosceles triangles be constructed upon the same base, their vertices will be all placed upon the right line, which is perpendicular to the base, and passes through its middle point. This is a very obvious and simple example of a species of theorem which frequently occurs in geometrical investigations. This perpendicular is said to be the locus of the vertex of isosceles triangles standing on the same base.

By this proposition it appears, that if the line G B makes the angle B G H equal to the supplement of G H D (84), the line G B will be parallel to H D. In the twelfth axiom (54) it is assumed, that if a line make an angle with G H less than the supplement of G H D, that line will not be parallel to H D, and will therefore meet it, if produced. The principle, therefore, which is really assumed is, that two right lines which intersect each other cannot be both parallel to the same right line, a principle which seems to be nearly self-evident.

If it be granted that the two right lines which make with the third, G H, angles less than two right angles be not parallel, it is plain that they must meet on that side of G H on which the angles are less than two right angles; for the line passing through G, which makes a less angle than B G H, with G H on the side B D, will make a greater angle than A G H with G H on the side A C; and therefore that part of the line which lies on the side A C will lie above A G, and therefore can never meet H C.

Various attempts have been made to supercede the necessity of assuming the twelfth axiom; but all that we have ever seen are attended with still greater objections. Neither does it seem ot us, that the principle which is really assumed as explained above can reasonably be objected against. See Appendix, II.

(117)   Cor. 1.—If two right lines which intersect each other (A B, C D) be parallel respectively to two others (E F, G H), the angles included by those lines will be equal.

Let the line I K be drawn joining the points of intersection. The angles C I K and I K H are equal, being alternate; and the angles A I K and I K F are equal, for the same reason. Taking the former from the latter, the angles A I C and H K F remain equal. It is evident that their supplements C I B and G K F are also equal.

(118)   Cor. 2.—If a line be perpendicular to one of two parallel lines, it will be also perpendicular to the other; for the alternate angles must be equal.

(119)   Cor. 3.—The parts of all perpendiculars to two parallel lines intercepted between them are equal.

For let A B be drawn. The angles B A C and A B D are equal, being alternate; and the angles B A D and A B C are equal, for the same reason; the side A B being common to the two triangles, the sides A C and B D must be equal (XXVI).

(120)   Cor. 4.—If two angles be equal (A B C and D E F), and the sides A B and D E be parallel, and the other sides B C and E F lie at the same side of them, they will also be parallel; for draw B E. Since A B and D E are parallel, the angles G B A and G E D are equal. But, by hypothesis, the angles A B C and D E F are equal; adding these to the former, the angles G B C and G E F are equal. Hence the lines B C and E F are parallel.

(122)   Cor.—Hence two parallels to the same line cannot pass through the same point. This is, in fact, equivalent to the twelfth axiom (115).

(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 called convex figures; that is, of figures in which every internal angle is less than two right angles. Some figures, however, have angles which are called reentrant angles, 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 a convex 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 2n - 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, if n be 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 angle is 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).

(152)   Cor. 1.—If two parallelograms have an angle in the one equal to an angle in the other, all the angles must be equal each to each. For the opposite angles are equal by this proposition, and the adjacent angles are equal, being their supplements.

(153)   Cor. 2.—If one angle of a parallelogram be right, all its angles are right; for the opposite angle is right by (151), and the adjacent angles are right, being the supplements of a right angle.

(154)   Both diagonals A D, B C being drawn, it may, with a few exceptions, be proved that a quadrilateral figure which has any two of the following properties will also have the others:

1° The parallelism of A B and C D.

2° The parallelism of A C and B D.

3° The equality of A B and C D.

4° The equality of A C and B D.

5° The equality of the angles A and D.

6° The equality of the angles B and C.

7° The bisection of A D by B C.

8° The bisection of B C by A D.

9° The bisection of the area by A D.

10° The bisection of the area by B C.

These ten data combined in pairs will give 45 distinct pairs; with each of these pairs it may be required to establish any of the eight other properties, and thus 360 questions respecting such quadrilaterals may be raised. These questions will furnish the student with a useful geometrical exercise. Some of the most remarkable cases are among the following corollaries:

The 9th and 10th data require the aid of subsequent propositions.

(155)   Cor. 3.—The diagonals of a parallelogram bisect each other.

For since the sides A C and B D are equal, and also the angles C A E and B D E, as well as A C E and D B E, the sides (XXVI) C E and B E, and also A E and E D are equal.

(156)   Cor. 4.—If the diagonals of a quadrilateral bisect each other, it will be a parallelogram.

For since A E and E C are respectively equal to D E and E B, and the angles A E C and D E B (XV) are also equal, the angles A C E and D B E are equal (IV); and, therefore, the lines A C and B D are parallel, and, in like manner, it may be proved that A B and C D are parallel.

(157)   Cor. 5.—In a right angled parallelogram the diagonals are equal.

For the adjacent angles A and B are equal, and the opposite sides A C and B D are equal, and the side A B is common to the two triangles C A B and A B D, and therefore (IV) the diagonals A D and C B are equal.

If the diagonals of a parallelogram be equal, it will be right angled.

For in that case the three sides of the triangle C A B are respectively equal to those of D B A, and therefore (VIII) the angles A and B are equal. But they are supplemental, and therefore each is a right angle.

^★★^★ (158)   The converses of the different parts of the 34th proposition are true, and may be established thus:

If the opposise sides of a quadrilateral be equal it is a parallelogram.

For draw A D. The sides of the triangles A C D and A B D are respectively equal, and therefore (VIII) the angles C A D and A D B are equal, and also the angles C D A and D A B. Hence the sides A C and B D, and also the sides A B and C D are parallel.

Hence the lozenge is a parallelogram, and a square has all its angles right.

If the opposite angles of a quadrilateral be equal, it will be a parallelogram.

For all the angles together are equal to four right angles (134); and since the opposite angles are equal, the adjacent angles are equal to half the sum of all the angles, that is, to two right angles, and therefore (XXVIII) the opposite sides are parallel.

If each of the diagonals bisect the quadrilateral, it will be a parallelogram.

This principle requires the aid of the 39th proposition to establish it. The triangles C A D C B D are equal, each being half of the whole area, therefore (XXXIX) the lines A B and C D are parallel. In the same manner D A B and D C B are equal, and therefore A C and B D are parallel.

^★★^★ (159)   The diagonals of a lozenge bisect its angles.

For each diagonal divides the lozenge into two isosceles triangles whose sides and angles are respectively equal.

^★★^★ (160)   If the diagonals of a quadrilateral bisect its angles, it will be a lozenge.

For each diagonal in that case divides the figure into two triangles, having a common base placed between equal angles, and therefore (VI) the conterminous sides of the figure are equal.

^★★^★ (161)   To divide a finite right line A L into any given number of equal parts.

From the extremity A draw any right line A X of indefinite length, and take upon it any part A B. Assume B C, C D, D E, &c. successively equal to A B (III), and continue in this manner until a number of parts be assumed on A X equal in number to the parts into which it is required to divide A L. Join the extremity of the last part E with the extremity L, and through B C D &c. draw parallels to E L. These parallels will divide A L into the required number of equal parts.

It is evident that the number of parts is the required number.

But these parts are also equal. For through b draw b m parallel to A E, and b c is a parallelogram; therefore b m is equal to B C or to A B. Also the angle A is equal to the angle c b m, and A b B to b c m. Hence (XXVI) A b and b c are equal. In like manner it may be proved, that b c and c d are equal, and so on.

(162)   Parallelograms whose sides and angles are equal are themselves equal. For the triangles into which they are divided by their diagonals have two sides and the included angles respectively equal, and are therefore (IV) equal, and therefore their doubles, the parallelograms, are equal.

(163)   Hence the squares of equal lines are equal.

(164)   Also equal squares have equal sides. For the diagonals being drawn, the right angled isosceles triangles into which they divide the squares are equal; the sides of those triangles must be equal, for if not let parts be cut off from the greater equal to the less, and their extremities being joined, an isosceles right angled triangle will be found equal to the isosceles right angled triangle whose base is the diagonal of the other square (IV), and therefore equal to half of the other square, and also equal to half of the square a part of which it is; thus a part of the half square is equal to the half square itself, which is absurd.

We have in this proof departed from Euclid in order to avoid the subdivision of the proposition into cases. The equality which is expressed in this and the succeeding propositions is merely equality of area, and not of sides or angles. The mere equality of area is expressed by Legendre by the word equivalent, while the term equal is reserved for equality in all respects. We have not thought this of sufficient importance however to justify any alteration in the text.

It is here supposed that the equal bases are placed in the same right line.

(167)   Cor.—If two opposite sides of a parallelogram be divided into the same number of equal parts, and the corresponding points of division be joined by right lines, these right lines will severally divide the parallelogram into as many equal parallelograms.