In either leg, as CA, produced, take any point D; on CB produced, take CE equal to CD [3. 1], and draw DB, AE. D A B In the triangles DCB, ECA, the sides CD, CE are equal [Constr.], also the sides CB, CA [Hyp.], and the angle C is common to both these triangles; therefore the angle CBD is equal to CAE, the angle CDB to CEA, and BD to AE [4. 1]. Whence, in the triangles ADB, BEA, the angle ADB is equal to BEA, the side DB to AE, and, taking the equals CA, CB, from the equals CD, CE, the side AD to BE [Ax. 3], therefore the angles DAB, ABE, which are the angles under the base, are equal [4. 1]. And, in the same triangles, the angles ABD, BAE are equal; which being taken from the equal angles CBD, CAE, the residues CBA, CAB, which are the angles at the base AB, of the triangle ABC, are equal [Ax. 3]. Corollary.-Hence every equilateral triangle is equiangular. For, whichever side be considered as base, the angles adjacent to it are equal, being opposite equal sides. PROP. VI. THEOR. If two angles (CAB, CBA) of a triangle (ABC) be equal, the sides (CA, CB), opposite to them, are equal. For, if CA and CB be not equal, let one of them, if possible, as CA, be the greater, and take from it AD equal to BC [3. 1], and draw BD. Because, in the triangles DAB, CBA, the sides DA, AB are severally equal to the sides CB, BA, and the angles DAB, CBA included A B by the equal sides, also equal [Hyp.]; the triangles DAB, CBA are themselves equal [4. 1], a part to the whole, which is absurd [Ax. 9], therefore the sides CA, CB are not unequal, they are therefore equal. Cor.-Hence every equiangular triangle is equilateral. For, whichever side be considered as base, the angles adjacent to it are equal, and therefore the sides opposite to them. PRÒP, VII. THEOR. Upon the same base (AB), and on the same side of it, there cannot be two triangles (ACB, ADB), whose conterminous sides are equal, (namely AC to AD, and BC to BD). For, if possible, let ACB, ADB be such; and first, let the vertex of each fall without the other. D Join CD, and because, in the triangle CAD, the sides AC, AD are equal [Hyp.], the angles ACD, ADC are equal [5. 1.]; but the angle ACD is greater than its part BCD [Ax. 9], therefore the angle ADC is greater than BCD; of course, the angle BDC, which is greater than ADC [Ax. 9], is greater than BCD: but, because the sides BC, BD of the triangle BDC are equal [Hyp.], the angles BDC, BCD are equal [5. 1.]; therefore the angle BDC is both equal to, and greater than, the angle BCD; which is absurd. Therefore two triangles on the same base and same side of it, have not their conterminous sides equal, if the vertex of each fall without the other. Let now, if possible, the vertex D of either triangle, as ADB, fall within the other. B Join CD, and produce AC, AD, as to E and F; and, because in the triangle CAD, the sides AC, AD are equal [Hyp.], the angles ECD, FDC on the other side of the base CD are equal [5. 1]; but the angle BDC is greater than FDC [Ax. 9], and therefore greater also than ECD; and the angle ECD is greater than BCD [Ax. 9]; therefore the angle BDC is greater than BCD: but, in the triangle BCD, because the sides BC, BD are equal [Hyp.], the angles BDC, BCD are equal [5. 1]; therefore the angle BDC is both equal to, and greater than, the angle BCD; which is absurd. Therefore two triangles, on the same base, and same side of it, have not their conterminous sides equal, if the vertex of either fall within the other. Lastly, let the vertex D, (see figure to the preceding prop.) of one, fall on one of the sides AC of the other; and in this case AC, AD are unequal [Ax. 9]. However therefore the vertices of the triangles fall, their conterminous sides are not equal. PROP, VIII. THEOR. If two triangles (ABC, DEF), have the two sides (CA, CB) of one, severally equal to the two sides (FD, FE) of the other, and have also their bases (AB, DE) equal; the vertical angles (C, F) are equal. For the triangle ABC being so applied to the triangle DEF, that the point A may coincide with the point D, and the right line AB with DE, the triangles ABC, DFE being to the same part, the point B would coin AA cide with E, because AB is equal to DE; and the whole triangle ACB would coincide with the triangle DFE, for if it should have a different position, as DGE, there would be on DE, and on the same side of it, two triangles, with equal conterminous sides, which is absurd [7. 1]; therefore the sides CA, CB would coincide with the sides FD, FE, and the angle C with the angle F, which angles are therefore equal [Ax. 8]. Corollary. Because CA, CB and the included angle C, are severally equal to FB, FE and the included angle F, the remaining angles A, B of the triangle ACB, are severally equal to the remaining angles FDE, FED of the triangle DFE, and also the triangles themselves. PROP. IX. PROB. To bisect, or divide into two equal parts, a given rectili- Take any point D in AB, and, on AC, take AE equal to AD [3. 1]; draw DE, on which make the equilateral triangle DFE [1. 1]; draw AF, which bisects the given angle BAC. For, in the triangles ADF, AEF, the sides AD, AE are equal [Constr.], AF common, and the bases DF, EF also equal [Constr.]; therefore the angles DAF, EAF are equal [8. 1], and of course the given angle BAC is bisected by the right line AF. A A B D E Cor. By the aid of this proposition, an angle may be divided into 2, 4, 8, 16, &c. equal parts, by repeatedly bisecting the several parts. PROP. X. PROB. To bisect a given finite right line (AB). On AB make the equilateral triangle ABC [1. 1], bisect the angle ACB by the right line CD, meeting AB in D [9. 1]. AB is bisected in D. For, in the triangles ACD, BCD, AC and BC are equal [Constr.], CD common, and the angles ACD, BCD also equal [Constr.], therefore the bases AD, DB are equal [4. 1], and so the right line AB is bisected in D. Α D PROP. XI. PROB. To draw a right line perpendicular to a given right line (AB), from a given point (C) therein. Take any point D in AC, on CB take CE equal to CD [3. 1], and on DE make the equilateral triangle DFE [1. 1]; draw CF, which is perpendicular to AB. For, in the triangles DCF, ECF, Athe sides DC, ČE, are equal -B [Constr.], CF common, and the bases DF, EF also equal [Constr.]; therefore the angles DCF, ECF opposite the equal bases are equal [8. 1], and so CF is perpendicular to AB [Def. 20]. Theorem. All right angles [as BAC, EDF,] are equal to each other. Let CA and FD be produced, as to G and H [Post. 2]. The angle BAC being applied to the angle EDF, so that the point A may coincide with the point D, and the right line AC with DF, the right line AG would coincide with DH, for if AG did not coincide with DH, but had a different situation, as DK, two right lines HD, KD would have a common segment DF, which is absurd [Ax. 11]; also AB would coincide with DE, for if not, let it, if possible, fall on either side of DE, as towards H, in the right line DL; then because EDF is equal to EDH (Def. 20), and LDF greater than EDF (Ax. 9), LDF is greater than EDH; whence, EDH being greater than LDH (Ax. 9), LDF is greater than LDH; but the angle BAC is equal to LDF (Hyp.), and BAG to LDH (Hyp.), therefore BAC is greater than BAG, which is absurd, BAC being, by supposition, a right angle, and therefore the angles BAC, BAG equal to each other (Def. 20). A like absurdity would follow, if AB were to fall on the other side of DE towards F. Therefore AB coincides with DE, and so the angles BAC, EDF coincide, and are of course equal [Ax. 8]. PROP. XII. PROB. From a given point (C), without a given right line (AB), producible at pleasure, to draw a perpendicular to it. Let any point whatever, as D, be taken on the other side of AB with respect to the given point C, and from the centre C, at the distance CD, let the circle EDF be described (Post. 3); which, because the points C, D are on different sides of the right line AB, meets the same right line, pro G B 10 duced if necessary, in two points, as E and F. Bisect EF in |