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CAB will be equal to the remaining angle CBA (Ax. 3.)
COROLLARY. Every equilateral triangle is also equiangular.
PRO P. VI. THEOREM.
If two angles of a triangle be equal to each other, the sides which are opposite to them will also be equal.
Let ABC be a triangle, having the angle CAB equal to the angle CBA; then will the fide ca be equal to the
For if ca be not equal to cb, one of them must be greater than the other; let ca be the greater, and make AD equal to bc (Prop. 3.), and join BD.
Then, because the two sides AD, AB, of the triangle ADB, are equal to the two sides BC, BA, of the triangle ACB, and the angle DAB is equal to the angle CBA (by Hyp.), the triangle ADB will be equal to the triangle ACB (Prop. 4.), the less to the greater, which is absurd.
The side ca, therefore, cannot be greater than the fide cs; and, in the same manner, it may be shewn that it cannot be less; confequently they are equal to each other. Q: E. D.
COROL. Every equiangular triangle is also equilateral,
If the three sides of one triangle be equal to the three sides of another, each to each, the angles which are opposite to the equal fides will also be equal.
Let ABC, DEF be two triangles, having the side AB. equal to the side DE, AC to Df, and sc to EF; then will the angle ACB be equal to the angle DFE, BAC to EDF, and ABC to def.
For, let the triangle Die be applied to the triangle Ace, so that their longest fides, DE, AB, may coincide, and the point F fall at G; and join cg.
Then, since the fide ac is equal to the side of, or AG (by Hyp.), the angle ACG will be equal to the angle AGC (Prop. 5.)
And, because the side bc is equal to the side EF, or BG (by Hyp.), the angle BCG will be equal to the angle BGC (Prop. 5.)
But since the angles ACG, BCG are equal to the angles AGC, BGC, each to each, the whole angle ACB will be equal to the whole angle AGB (Ax. 8.)
And, because ac is equal to AG, BC to BG, and the angle ACB to the angle AGB, the angle CAB will, also,
be equal to the angle GAB, and the angle ABC to the angle ABG (Prop. 4.).
But the triangles AGB, DFE, are identical; consequently the angles of the triangle Dfe will, also, be equal to the corresponding angles of the triangle ACB. Q. E. D.
Let ABC, DEF be each of them right angles; then will abc be equal to def.
For conceive the angle der to be applied to the angle ABC, so that the point E may coincide with the point B, and the line ed with the line ba.
And if EF does not coincide with bc, let it fall, if polible, without the angle ABC, in the direction BG; and produce AB to H.
Then, because the angles ABC, ABG are right angles (by Hyp.), the lines CB, GB will be each perpendicular to AH (Def. 8. 9.)
And, since a right line which is perpendicular to another right line, makes the angles on each side of it equal (Def. 8.), the angle CBA will be equal to the angle CBH, and the angle GBA to the angle GBH.
But the angle gba is greater than the angle CBA, or its equal CBH; consequently thc angle GBH is also greater
than the angle CBH; that is, a part is greater than the whole, which is absurd.
The line EF, therefore, does not fall without the angle ABC; and in the same manner it may be shewn that it does not fall within it; consequently EF and bc will cos incide, and the angle der be equal to the angle ABC, as was to be shewn.
PRO P. IX. PROBLEM.
To bisect a given rectilineal angle, that is, to divide it into two equal parts.
Let Bac be the given rectilineal angle ; it is required to divide it into two equal parts.
Take any point D in AB, and from Ac cut off AE equal to AD (Prop. 3.), and join de.
Upon de describe the equilateral triangle DFE (Prop. 1.), and join AF; then will AF bisect the angle BAC, as was required.
For AD is equal to AE, by construction; DF is also equal to FE (Dif. 16.), and AF is common to each of the triangles AFD, AFE.
But when the three sides of one triangle are equal to the three sides of another, each to each, the angles which are opposite to the equal sides are, also, equal (Prop. 7.)
The side DF, therefore, being equal to the side Fe, the angle day will be equal to the angle Fae; and confequently the angle bac is bisected by the right line AF, as was to be done.
PROP. X. PROBLEM.
To bisect a given finite right line, that is, to divide it into two equal parts.
Let Ac be the given right line; it is required to divide it into two equal parts.
Upon Ac describe the equilateral triangle ACB (Prop. 1.), and bife&t the angle ABC by the right line BD (Prop. 9.); then will ac be divided into two equal parts at the point D, as was required.
For AB is equal to Bc (Def. 16.), bp is common to each of the triangles ADB, CDB, and the angle ABD is equal to the angle CBD (by Const.)
But when two sides and the included angle of one triangie, are equal to two sides and the included angle of another, each to each, their bases will also be equal (Prop. 4.)
The bafe ad is, therefore, equal to the base Dc; and, confequently, the right line Ac is bisected in the point D, as was to be done.