Book I. Because the point B is the centre of the circle CGH, BC is equal to BG; and because D is the centre of the * 15 Def. circle GKL, DL is equal to DG, and DA, DB, parts of them, are equal; therefore the remainder AL is equal to €3 Ax. the remainderf BG; But it has been shown, that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the same are equal to one another; therefore the straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. Which was to be done, * 2. 1. PROP. III. PROB. Let AB and C be the two given C E B straight line AD equal to C; and from the centre A, and at the di63 Post. stance AD, describeb the circle F DEF; and because A is the centre of the circle DEF, AE shall be equal to AD; but the straight line C is likewise equal to AD; whence AE and C are each of them equal to AD; wherefore the straight "1 Ax. line AE is equal toc C, and from AB, the greater of two straight tines, a part AE has been cut off equal to C the less. Which was to be done. PROP. IV. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each ; and have likewise the angles contained by those sides equal to one another; they shall likewise have their bases, or third sides, equal; and the two triangles shall be equal; and their other angles shall be equal, each to each, viz.those to winch the equal sides are opposite. Let ABC, DEF be two triangles, which have the twosides AB, AC equal to the two sides DE, DF. each to each, viz. AB to DE, and AC to Book I. DF;and the angle BAC А D 3. C E For, if the triangle ABC be applied to DEF, so that the Pato PROP. V. THEOR. Le, ABC be an isosceles triangle, of which the side AB is Prvox I. equal to AC, and let the straight lines AB, AC be produced to D and E, the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE. In BD take any point F, and from AE the greater, cut * 5. 1. off AG equal to AF, the less, and join FC, GB. Because AF is equal to AG, and AB to AC, the two sides FA, AC are equal to the two GA, AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is * 4.1. equal to the base GB, and the tri angle AFC to the triangle AGB; G E whole AF is equal to the whole AG, of which the parts AB, AC, are equal; the remainder •3 Ax. BF shall be equal to the remainder CG; and FC was proved to be equal to GB; therefore the two sides BF, FC are equal to the two CG, GB, each to each; and the angle BFC is equal to the angle CGB, and the base Be is common to the two triangles BFC, CGB; wherefore the triangles are equalb, and their 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: And, since it has been demonstrated, that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are also equal; the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC: And it has also been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore the angles at the base, &c. Q.E.D. COROLLARY. Hence every equilateral triangle is also equiangular. PROP. VI. THEOR. If two angles of a triangle be equal to one another, the sides also which subtendx or are opposite to, the equal angles, shall be equal to one another. pl Let ABC be a triangle having the angle ABC equal to Boox the angle ACB; the side AB is also equal to the side AC. For, if AB be not equal to AC, one of them is greater than the other: Let AB be the greater ; and from it cuta a 3. 1. off DB equal to AC, the less, and join DC; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both the two sides, A DB, BC are equal to the two AC, CB each to each; and the angle DBC is D equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangleb ACB, the less to the greater; which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it Wherefore, if two angles, &c. Q. E. D.B Cor. Hence every equiangular triangle is also equilateral. b 4.1. PROP. VII. THEOR. UPON the same base, and on the same side of See N. it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. If it be possible, let there be two triangles ACB, ADB, upon the same base AB, and upon the same side of it, which have their sides CA, DA terminated in the extremity A of the base equal to one another, and, likewise their sides, CB, DB, that are terminated in B. Join CD; then, in the case in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equala to the a 5. 1. angle ADC: But the angle ACD is greater than the angle BCD; А" therefore the angle ADC is greater also than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equals to the angle BCD; but it has been demonstrated to be greater than it; which is impossible. : Book I. But if one of the vertices, as D, be within the other trimangle ACB; produce AC, AD to E, F; therefore, because AC is equal to AD in the triangle E F the other side of the base CD are gle ECD is greater than the angle A the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. Q. E. D. PROP. VIII. THEOR. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them, of the other. Let ABC, DEF be two triangles, having the two sides AB, AC, equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF; and also the base BC equal to the A D G For, if the E |