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The first case is one of the two in which the triangle is not determined; for a triangle may have its sides increased or diminished to any extent without altering the magnitude of its angles.

The second case is demonstrated in the eighth proposition.

The third case is the subject of the present proposition.

B

The fourth case is the other one in which the triangle is not determined; for it is quite possible to have two triangles having two sides of the one equal to two sides of the other, and one of the opposite angles of the one equal to the similar angle of the other, and yet for the triangles themselves not to be equal. Thus, let ABC be a triangle in which neither A nor C are right angles and A is less than C; then from B as a center, and the distance BC as radius, describe a circle cutting AC in D, and draw DB. Now it is evident that, in the triangles ABC and ABD, we have the two sides AB and BC equal to the two AB and BD, and the opposite angle A the same in both; and yet the two triangles are not equal.

The fifth and sixth cases are demonstrated in the twenty-sixth proposition. 3. The application of one figure to another, so as to prove or disprove their coincidence, as made use of in this proposition, is termed superposition. It has been objected to by some mathematicians as not being strictly geometrical; but, as Mr. De Morgan has observed, it requires only the admission of the following postulate, "That any figure may be removed from place to place without alteration of form, and a plane figure may be turned over on the plane." The latter portion, printed in italics, is required for the proof of the fifth proposition.

4. The enunciation of this proposition really includes three distinct propositions, which have been distinguished by separate numbers. The demonstration of the first of these is an example of the negative or indirect proof termed "Reductio ad Absurdum," which consists in proving a proposition by showing that if it is denied an obvious absurdity follows. Concerning this method of proof, see the Introduction.

5. The base of a triangle is the third side of a triangle as distinguished from the other two, without any regard to whether the triangle stands upon

it or not.

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PROPOSITION V.

THEOREM. [1] If a triangle (ABC) be isosceles, the angles at the base (ABC and ACB) are equal to one another; [2] and if the equal sides be produced, the angles formed by the produced sides and the base below the same (CBD and BCE) shall be equal.

CONSTRUCTION. Produce the equal sides AB and AC (a), and in the produced part of one of them AB take any point F, and from the other cut off AG equal to AF (b). Draw a straight line from Ĉ to F, and from B to G (c).

D

B

A

(a) Post. 2.
(b) I. 3.
(c) Fost. 1.

DEMONSTRATION. [2.] In the triangles ACF and ABG, the side AC is equal to the side AB (d), the side AF to the side AG (e), and the angle A is common to both; therefore the base CF is equal to the base BG, the angle ACF to the angle ABG, and the angle F to the angle G (f); then taking the equal lines AC and AB from the equal lines AG and AF, the remainders CG and BF are equal (g); therefore in the triangles BCG and CBF, because the side CG is equal to the side BF, the side BG to the side CF, and the angle G to the angle F, the angle BCG must be equal to the angle CBF (f), which are the angles formed by the produced sides and the base, below the same.

(d) Hypoth. (e) Constr. (ƒ) I. 4. (9) Ax. 3.

[1.] Further in the same triangles the remaining angles must be equal, BCF to CBG (f); and if these equal angles be taken from the equal angles ACF and ABG, the remaining angles, ACB and ABC, will be equal (g), which are the angles at the base of the given triangle.

A

COROLLARY. Hence every equilateral triangle is also equiangular; for if each side be taken in succession as the base, it may be shown that the angles adjacent to the side so taken are equal. SCHOLIA. 1. This proposition may also be proved in the following manner: if the triangle ABC be turned over on the plane (see scholium 3 to the preceding proposition), so that the position of the point A may be unaltered, while the side AB lies on AC, then, since the angle A is the same in both, the side AC must fall on AB; and because the sides AB and AC are equal (a), the point C will coincide with the point B, the point B with C, the angle ACB with the angle ABC, and the angle BCG with the angle CBF; and therefore [1] the angle ACB will be equal to the angle ABC (b), and [2] the angle BCG equal to the angle CBF (b).

2. The enunciation of this proposition really includes two separate propositions which have been distinguished by numbers.

PROPOSITION VI.

D

F

B

(a) Hypoth. Ax. 8.

A

THEOREM.-If two angles (B and C) of a triangle (ABC) are equal, the sides (AC and AB) opposite to those angles are also equal.

DEMONSTRATION. For if AB be not equal to AC, one of them is greater than the other; let AB be the greater, from it cut off DB equal to AC the less (a), and draw the line DC. Then in the triangles ABC and DBC, because the side AC is equal to the side DB, the base BC common to both, and the angle ACB equal to the angle DBC (b), therefore the tri

B

(a) I. 3.
(b) Hypoth.
(c) I. 4.

angles themselves are equal (c); the greater ABC to the lesser DBC, which is absurd; therefore neither of the sides AC or AB being greater than the other, they are equal.

COROLLARY. Hence every equiangular triangle is also equilateral, which may be shown by taking each side in succession as the base.

SCHOLIA. 1. It should be observed that the portion equal to the lesser side must be cut off from that end of AB next to the equal angle; otherwise no proof can be drawn from proposition iv.

2. This proposition is the converse of the first part of the preceding; that is to say, the hypothesis of one is the predicate in the other, and vice versû, which will be immediately seen by expressing them as under:PROP. V. If two sides are equal, the opposite angles are equal. PROP. VI. If two angles are equal, the opposite sides are equal.

The truth of a proposition does not establish that of its converse, for the reasons explained in treating of the conversion of propositions in the Introduction.

This proposition is demonstrated by the method of "reductio ad absurdum," which will be found, in the Elements, to be most frequently employed in the demonstration of converse propositions.

PROPOSITION VII.

THEOREM.-If two triangles (ABC and ABD) be upon the same base (AB) and on the same side of it, they cannot have their sides which are terminated in one extremity of that base equal to one another, and also those which are terminated in the other extremity (AC to AD and BC to BD).

DEMONSTRATION. If it be possible, let there be two triangles on the line AB; then must either [1] the vertex of each of the triangles be without the other one, or [2] the vertex of one triangle within the other one, or [3] upon one side of it.

A

[1] Let the vertex of each triangle be without the other one. Draw a straight line from C to D, the two vertices. Then because in the triangle BCD the sides BC and BD are equal (a), therefore the angles BDC and BCD are equal (b). Also, because in the triangle ACD the sides AC and AD are equal (a), therefore the angles ADC and ACD are equal (b). Now the angle ACD is greater than BCD (c); and because ACD and ADC are equal, therefore is ADC also greater than BCD; and as BDC is greater than ADC (c), therefore BDC is greater than BCD; but they have already been proved to be equal, which is absurd.

(a) Hypoth. 1.5.

Ax. 9.

[2.] Let the vertex of one triangle be within the other one. Produce the sides AC and AD, and draw a line from C to D. Then, because in the triangle BCD the sides BC and BD are equal (a), therefore the angles BDC and BCD are equal (b). Also, because in the triangle ACD the two sides AC and AD are equal (a), therefore the angles ECD and FDC, on the other side of the base, are equal (b). Now the angle ECD is greater than the angle BCD (c); and because FDC and ECD are equal, therefore is FDC also greater than BCD; and as BDC is greater than FDC (c), therefore is BDC greater than BCD; but they have already been proved to be equal, which is absurd.

[3.] Let the vertex D of one triangle fall on the side AC of the other. Then it is evident that the sides BC and BD are not equal, which is contrary to the hypothesis.

Therefore in no case can two triangles be upon the same base, and upon the same side of it, that have their sides which are terminated in one extremity of that base equal to one another, and also those which are terminated in the other extremity.

E

(a) Hypoth.

(b) I. 5.
(c) Ax. 9.

SCHOLIA. 1. The only use made of this proposition is to prove that which follows it, which can, however, be demonstrated without it, as is done in the scholium attached to that proposition.

2. The form of argument adopted by Euclid in the demonstration of this proposition is that termed a dilemma, and this is the only instance throughout the Elements in which it is employed. The argument will also be seen to be by the "reductio ad absurdum."

PROPOSITION VIII.

THEOREM.-If two triangles (ABC and DEF) have two sides of the one respectively equal to two sides of the other (AB and AC to DE and DF), and have also their bases equal (BC to EF), then the angles (A and D) formed by the equal sides are equal.

DEMONSTRATION. For if the triangle ABC be applied to the triangle DEF, so that their equal bases may coincide, and that the two triangles may lie on the same side, their equal sides must coincide (a), BA with ÉD, and CA with FD, and therefore the angles A and D must coincide, and be equal to each other (b).

A

E

(a) I. 7.
(b) Ax. 8.

COROLLARY. Hence also the angles opposite the equal sides are equal, B to E and C to F; and also the triangles themselves are equal.

(a) Hypoth. I. 5.

SCHOLIA. 1. This proposition may be demonstrated in the following manner, without any reference to the seventh. Let the triangle ABC be applied to the triangle DEF, so that their bases may coincide, and that the two triangles may lie on opposite sides; join the vertices DG. Then, because in the triangle DEG the two sides ED and EG are equal (a), therefore the angles EDG and EGD are equal (b). Also, because in the triangle DFG the two sides FD and FG are equal (a), therefore the angles FDG and FGD are equal (b). Then if the equal angles EDG and EGD be added to the equal angles FDG and FGD, the whole angles EDF and EGF will be equal (c). But the angle EGF is equal to BAC (a), therefore the angles EDF and BAC are equal (d). 2. This proposition is the converse of proposition iv. When a theorem has several hypotheses and one predicate, if another theorem be framed having one of those hypotheses for its predicate and the predicate of the first as one of its hypotheses, the two theorems are the converse of each other; and it is in this sense that the eighth proposition is the converse of the fourth, as will be immediately seen by expressing them in the following

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(c) Ax. 2.
(d) Ax. 1.

Sthe angles opposite then the bases are equal.
the bases are equal, S

the angles opposite

PROP. VIII.-If two sides are equal then the bases are equal. and the bases are equal,

PROPOSITION IX.

PROBLEM.-To bisect a given rectilineal angle (BAC).

SOLUTION. Take any point D in AB, and from AC cut off AE equal to AD (a); draw DE, and upon the side furthest from A construct an equilateral triangle DEF (b); then a straight line drawn from A to F will bisect the angle BAC.

DEMONSTRATION. Because in the triangles AFD and AFE, the sides AD and AE are equal (c), the side AF common to both, and the base DF equal to the base EF (c); therefore the angle DAF is equal to the angle EAF (d), and the given rectilineal angle BAC is bisected by the straight line AF.

B/

(a) I. 3.

(b) I. 1.

(c) Solution.
(d) I. 8.

SCHOLIA. 1. The direction to construct the equilateral triangle on the side of DE furthest from A is given in order to avoid the possibility of the vertex falling on the point A, in which case the line AF could not be drawn. 2. By a repetition of this problem, an angle may be divided into

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