Demonstration.

For A E and AD are equal, being radii of the same circle (17); and C and A D are equal by the construction. Hence A E and C are equal.

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.

(62)

If two triangles (BAC and ED F) have two sides (BA and A C) in the one respectively equal to two sides (E D and DF) in the other, and the angles (A and D) included by those sides also equal; the bases or remaining sides (B C and EF) will be equal, also the angles (B and C) at the base of the one will be respectively equal to those (E and F) at the base of the other which are opposed to the equal sides (i. e. B to E and C to F).

Да

Let the two triangles be conceived to be so placed that the vertex of one of the equal angles D shall fall upon that of the other A, that one of the sides DE containing the given equal angles shall fall upon the side A B in the other triangle to which it is equal, and that the remaining pair of equal sides A C and D F shall lie at the same side of those A B and D E which coincide.

B

C

E

Since then the vertices A and D coincide, and also the equal sides A B and D E, the points B and E must coincide. (If they did not the sides A B and D E would not be equal.) Also, since the side D E falls on A B, and the sides A C and D F are at the same side of A B, and the angles A and D are equal, the side D F must fall upon AC; (for otherwise the angles A and D would not be equal.)

Since the side D F falls on A C, and they are equal, the extremity F must fall on C. Since the extremities of the bases B C and EF coincide, these lines themselves must coincide, for if they did not they would include a space (52). Hence the sides B C and E F are equal (50).

Also, since the sides E D and E F coincide respectively with B A and B C, the angles E and B are equal (50), and for a similar reason the angles F and C are equal.

C

Since the three sides of the one triangle coincide respectively with the three sides of the other, the triangles themselves coincide, and are therefore equal (50).

6

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 upon 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 DF 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:

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

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

3o 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 may 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.

PROPOSITION V. THEOREM.

(63) The angles (B, C) opposed to the equal sides (A C and A B) of an isosceles triangle are equal, and if the equal sides be produced through the extremities (B and C) of the third side, the angles (D B C and E C B) formed by their produced parts and the third side are equal.

Let the equal sides A B and A C be produced through the extremities B, C of the third side, and in the produced part B D of either let any point F be assumed, and from the other let A G be cut off equal to A F (III). Let the points F and G so taken on the produced sides be connected by right lines F C and B G with the alternate extremities of the third side of the triangle.

F

B

D

A

In the triangles F A C and G A B the sides FA and A C are respectively equal to GA and A B, and the included angle A is common to both triangles. Hence (IV), the line FC is equal to B G, the angle AF C to the angle A G B, and the angle A C F to the angle A B G. If from the equal lines A F and A G, the equal sides A B and A C be taken, the remainders B F and C G will be equal. Hence, in the triangles B F C and C G B, the sides B F and F C are respectively equal to CG and G B, and the angles F and G included by those sides are also equal. Hence (IV), the angles FBC and G C B, which are those included by the third side B C and the productions of the equal sides A B and A C, are equal. Also, the angles F C B and GBC are equal. If these equals be taken from the angles F CA and G B A, before proved equal, the remainders, which are the angles A B C and A C B opposed to the equal sides, will be equal.

(64) COR.-Hence, in an equilateral triangle the three angles are equal; for by this proposition the angles opposed to every two equal sides are equal.

(65) If two angles (B and C) of a triangle (BAC) be equal, the sides (A C and A B) opposed to them are also equal.

For if the sides be not equal, let one of them A B be greater than the other, and from it cut off D B equal to A C (III), and draw C D.

D

Then in the triangles DB C and ACB, the sides DB and BC are equal to the sides A C and CB respectively, and the angles D B C and AC B are also equal; therefore (IV) the triangles themselves D B C and ACB are equal, a part equal to the whole, which is absurd; therefore neither of the sides A B or A C is greater than the other; they are therefore equal to one another.

(66) COR.-Hence every equiangular triangle is also equilateral, for the sides opposed to every two equal angles are equal.

Since the three sides of the one triangle coincide respectively with the three sides of the other, the triangles themselves coincide, and are therefore equal (50).

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 upon 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 DF 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:

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

2o 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 may 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.

PROPOSITION V. THEOREM.

(63) The angles (B, C) opposed to the equal sides (A C and A B) of an isosceles triangle are equal, and if the equal sides be produced through the extremities (B and C) of the third side, the angles (D B C and EC B) formed by their produced parts and the third side are equal.

Let the equal sides A B and A C be produced through the extremities B, C of the third side, and in the produced part B D of either let any point F be assumed, and from the other let A G be cut off equal to AF (III). Let the points F and G so taken on the produced sides be connected by right lines F C and B G with the alternate extremities of the third side of the triangle.

In the triangles F A C and G A B the sides FA and A C are respectively equal to G A and A B, and the included angle A is common to both triangles. Hence (IV), the line FC is equal to B G, the angle A F C to the angle A G B, and the angle A C F to the angle A B G. If from the equal lines A F and A G, the equal sides A B and A C be taken, the remainders BF and C G will be equal. Hence, in the triangles B F C and C G B, the sides B F and F C are respectively equal to CG and G B, and the angles F and G included by those sides are also equal. Hence (IV), the angles FBC and G C B, which are those included by the third side B C and the productions of the equal sides A B and A C, are equal. Also, the angles F C B and G B C are equal. If these equals be taken from the angles F CA and G B A, before proved equal, the remainders, which are the angles A B C and A C B opposed to the equal sides, will be equal.

(64) COR.-Hence, in an equilateral triangle the three angles are equal; for by this proposition the angles opposed to every two equal sides are equal.

(65) If two angles (B and C) of a triangle (BAC) be equal, the sides (A C and A B) opposed to them are also equal.

For if the sides be not equal, let one of them A B be greater than the other, and from it cut off D B equal to A C (III), and draw C D.

B

D

A

Then in the triangles DB C and ACB, the sides DB and BC are equal to the sides A C and C B respectively, and the angles D B C and AC B are also equal; therefore (IV) the triangles themselves D B C and AC B are equal, a part equal to the whole, which is absurd; therefore neither of the sides A B or A C is greater than the other; they are therefore equal to one another.

(66) COR.-Hence every equiangular triangle is also equilateral, for the sides opposed to every two equal angles are equal.