one triangle equal to the other triangle; also the remaining angles of the one shall be equal to the remaining angles of the other, each to each, which are opposite to the equal sides.

Let there be two triangles, ABC, DEF, which have the two sides ab, ac, equal to the two sides de, df, each to each; namely, the side AB equal to the side DE, and the side Ac equal to DF; also the angle BAC equal to the angle EDF. Then is the base

AA

BC equal to the base EF, and the
triangle ABC equal to the triangle
DEF; also the remaining angles
equal to the remaining angles, R
each to each, to which the equal

CE

sides are opposite; namely, the angle ABC to the angle DEF; and the angle ACB to the angle DFE.

For if the triangle ABC be applied to the triangle DEF, and the point a be put upon the point D, and the right line AB upon the right line DE, then shall the point B coincide with the point E, because AB is equal to DE. But AB coinciding with DE; the right line Ac shall also coincide with the right line DF, since the angle BAC is equal to the angle EDF. Wherefore c will also coincide with F: for the right line AC is equal to the right line DF; but the point в coincides with the point E. Therefore the base BC will also coincide with the base EF. Because if the point в coinciding with the point E, and c with F; the base BC does not coincide with the base EF; two right lines would inclose a space which is impossible. Whence the base BC Ax. 10. coincides with the base EF, and also equal to it. Therefore the whole triangle ABC will coincide with the whole triangle DEF, and will be equal to it; also the remaining angles will coincide with the remaining angles, and equal to them, viz., the angle ABC to the angle DEF, ↳ Ax. 8. and the angle ACB to the angle DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, &c. Q. E. D.

a 4.1.

PROPOSITION V.

THEOREM.*

The angles which are at the base of isosceles triangles are equal to one another; and the equal right lines being produced, the angles under the base shall be equal to one another.

F

B

C

G

There

Let ABC be an isosceles triangle, having the side AB equal to the side Ac, and produce the right lines Ab, AC, directly forward to D, E. Then is the angle ABC equal to the angle ACB, and the angle CBD to the angle BCE. For take in the line BD any point F: and from the greater AE cut off AG equal to AF the less: also join FC, GB. Therefore, because AF is equal to AG; and AB to AC; the two D FA, AC, are equal to the two GA, AB, each to each; and contain the common angle FAG. fore the base rc is equal to the base GB, and the triangle AFC equal to the triangle AGB also the remaining angles shall be equal to the remaining angles, each to each, viz., the angle ACF equal to the angle ABG; also the angle AFC to the angle AGB. And because the whole AF is equal to the whole AG; of which the parts AB, AC, are equal; the remaining part BF will also be equal to the remaining part CG. But it has been proved that Fc is equal to GB. Therefore the two BF, FC, are equal to the two CG, GB, each to each; and the angle BFC equal to the angle CGB: also their base BC is common. Whence the triangle BFC is equal to the triangle CGB; and the remaining angles equal to the remaining angles, each to each, to which the equal sides are opposite. Therefore the angle FBC is equal to the angle GCB; and the angle BCF to the angle CBG. Wherefore, because the whole angle ABG has been proved to be equal to the

FC

* This theorem was discovered by Thales, for he is first said to have perceived and proved, that the angles at the base of every isosceles triangle are equal, and, after the manner of the ancients, to have called them similar. The latter part of it is not at all necessary in demonstrating the former; and it is affirmed by some geometers, amongst whom is Scarborough, that it is not Euclid's, but added by some one else; however this may be, the angles opposite the equal sides may be demonstrated without proving the equality of the angles under the base, as is evident by the very elegant and concise demonstration of Pappus, and indeed by many others.

b

whole angle ACF, of which the angle CBG is equal the angle BCF; the remaining angle ABC will be equal to Ax. 3. the remaining angle ACB: and they are at the base of the triangle ABC. But it has also been proved, that the angle FBC is equal to the angle GCB, which are under the base. Therefore the angles which are at the base of isosceles triangles are equal to one another; and the equal right lines being produced, the angles under the base shall be equal to one another. Q. E. D.

COROLLARY.

Hence every equilateral triangle is also equiangular.

PROPOSITION VI.

THEOREM.

If two angles of a triangle be equal to one another, the sides subtending the equal angles shall be equal to one another.

a 3. 1.

Let ABC be a triangle, having the angle ABC equal to the angle ACB. Then is the side AB equal to the side AC. For if AB be unequal to ac, one of them is greater. Let AB be the greater; and from the greater AB take away DB equal to AC the less; and join DC. join DC. Therefore because DB is equal to AC; and BC common, the two DB, BC, will be equal to the two AC, CB, each to each; and the angle DBC equal to the angle ACB (by hypoth.). Whence the base Dc is equal to 4.1. to the base AB, and the triangle DBC equal to the triangle ACB, the less to the greater, which is absurd. Therefore the sides AB, AC, are not unequal. Whence they are equal. Wherefore if two angles of a triangle be equal to one another, &c. Q. E. D.

COROLLARY.

DC

B

b

Hence every equiangular triangle is also equilateral.

PROPOSITION VII.
THEOREM.

On the same right line cannot be constituted two right lines equal to two other right lines, each to each, drawn to

a 5. 1.

b 5. 1.

different points, to the same parts, and having the same extremes with the two right lines first drawn.

For if it be possible, let the two right lines AD, Db, be constituted upon the right line AB equal to two right lines AC, CB, each to each, drawn to different points, c and D, situated on the same side of the line AB, the lines AD, DB, having the same ends A, B, with the two first lines AC, CB; so that CA be equal to DA, both having the same end A; and CB be equal to DB, both having the same end B: for join the right line CD. Therefore because AC is equal to AD, the angle ACD will be equal to the angle ADC. Whence the angle ADC is greater than the angle Wherefore the angle BDC will be much greater than the angle BCD. Again, because CB is equal to DB, the angle BDC will be equal to the angle BCD.

BCD.

But it

C D

A

B

has been shown to be much greater, which is impossible. Therefore on the same right line cannot be constituted two right lines, &c.

SCHOLIUM.

Q. E. D.

C

E/

F

If D, one of the points c, D, be within the triangle ACB, a demonstration may be obtained by means of the latter part of the fifth proposition. For AC, AD, being drawn, the external angles ECD, FDC, which are under the base of the isosceles triangle ACD, will be equal to one another; therefore the angle BDC will be greater than the angle ECD; whence the angle BDC will be much greater than the angle BCD; but because BD, BC, are equal, the angle BDC will be equal to the angle BCD, a greater to a less, which is impossible.

A

But if the point D be taken in either of them, AC, bc, the proposition is manifest; for so the whole ac would be equal to its part AD, or the whole BC equal to its part BD, which is impossible.

Dr. Simson, in his note to this proposition, says, he has thought proper to change its enunciation, "because (he adds) the literal translation from the Greek is extremely harsh, and difficult to be understood by beginners." Whatever difficulty learners may experience in this proposition, considered abstractedly, is easily removed by its exposition in the figure; and therefore it appears to me, that Dr. Simson has acted very injudiciously in altering its

enunciation: and I perfectly agree with Taylor in saying, that it seems strange such liberties should be taken by one, who professes, in his preface, to remove blemishes and restore the principal books of the Elements to their original accuracy.

PROPOSITION VIII.

THEOREM.

If two triangles have two sides equal to two sides, each to each, and have their bases equal; the angle also, which is contained by the equal sides of the one triangle, shall be equal to the angle contained by the equal sides of the other.

Let there be two triangles ABC, DEF, which have two sides AB, AC, equal to two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF; and they have the base BC equal to the base EF. Then is the angle BAC equal to the angle EDF. For the triangle ABC being applied to the triangle DEF, and the point B being put on E; also the right line BC being applied to EF, the point c will coincide with

B

CE

F

the point F, because BC is equal to EF. Therefore BC coinciding with EF; BA, AC, will also coincide with ED, DF; for if the base BC coincide with the base EF; and the sides BA, AC, do not coincide with the sides ED, DF, but have a different situation, as EG, GF; then, on the same right line would be constituted two right lines equal to two other right lines, each to each, drawn to different points, to the same parts, and having the same extremes with the two right lines first drawn. But they cannot be so constituted as has been demonstrated.a a 7. 1. Therefore if the base BC coincide with the base EF, the sides BA, AC, cannot but coincide with the sides ED, DF. Wherefore the angle BAC will also coincide with the angle EDF, and be equal to it. Therefore if two triangles have two sides equal to two sides, &c. Q. E. D.

Deduction from Euclid.

In an isosceles triangle, the right line drawn from the vertical angle bisecting the base is at right angles to the base.