Book I. IX. X. XI. XII. “ two interior angles on the fame fide of it taken together less “ than two right angles, these straight lines being continually “ produced shall at length meet upon that side on which are “ the angles which are less than two right angles. See the “ notes on Prop. 29. of Book I." 1 PROPOSITION I. PROBLEM. Book I. To defcribe an equilateral triangle upon a given in late. finite straight line. Let AB be the given straight line, it is required to describe an equilateral triangle upon it. From the center A, at the diAtance AB describe the circle a. 3d Portu. BCD. and from the center B, at the distance BA describe the circle D A В Е ACE; and from the point C in which the circles cut one another draw the straight lines CA, CB b. ad Polt. to the points A, B. ABC shall be an equilateral triangle. Because the point A is the center of the circle BCD, AC is equal to AB. and because the point B is the center of the circle ACE, c. 15th DeBC is equal to BA. but is has been proved that CA is equal to finition. AB; therefore CA, CB are each of them equal to AB. but things which are equal to the fame are equal to one another d; therefore d. st Axi. CA is equal to CB. wherefore CA, AB, BC are equal to one another. and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB. Which was required to be done. om. FRON PRO P. II. PROB. to a given straight line. From the point A to B draw the straight line AB ; and upon it de K a. I Port. scribe bthe equilateral triangle DAB, b. i. 1. and produce the straight lines DA, DB to E and F; from the center B, at the distance BC described the c d. 3. Poft, circle CGH, and from the center D, B at the distance DG describe the circle GKL. AL shall be cqual to F BC. H C. 2. Poft. Book 1, Because the point B is the center of the circle CGH, BC is eW qual e to BG. and because D is the center of the circle GKL,DL e.is. Def. is equal to DG, and DA, DB parts of them are equal; therefore f. 3. Ax. the remainder AL is equal to the remainder f BG. but it has been fnewn that BC is equal to BG; wherefore AL and BC are each of them equal to BG. and things that are equal to the fame 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. FR the greater. PRO P. III. PRO B. off a part equal to the less. It is required to cut E B From the point A draw the straight line AD equal to C; and from the center A, and at the di F b. 3. Port. {tance AD describe the circle DEF. and because A is the center 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 C. 1. Ax. them equal to AD. wherefore the straight line AE is equal to C, and from AB the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to be done. PROP. IV. THEORE M. two sides of the other, each to each; and have likewise the angles contained by those sides equal to one another : they shall likewise bave 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 which the equal fides are opposite. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF; and the an. A D Book I. gle BAC equal to the angle EDF. the base BC shall be eequal to the base EF; and the triangle ABC to the triangle DEF; and the other angies, to which the equal sides are opposite, shall be equal, each B СЕ. to each, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. For if the triangle ABC be applied to DEF so that the point A may be on D, and the straight line AB upon DE; the point B shall coincide with the point E, because AB is equal to DE. and AB coin. ciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF. wherefore also the point C shall coincide with the point F, because the straight line AC is equal to DF. but the point B coincides with the point E ; wherefore the base BC shall coincide with the base EF. because the point B coinciding with E, and C with F, if the base BC does not coincide with the base EF, two straight lines would inclose a space, which is impoflible". Therefore the base BC shall coincide with the a, 10. Ax. bafe EF, and be equal to it. Wherefore the whole triangle ABC fhall coincide with the whole triangle DEF, and be equal to it; and the other angles of the one shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore 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; their bases shall likewise be equal, and the triangles be equal, and their other angles to which the equal sides are opposite, shall be equal, each to each. Which was to be demonstrated. PROP. V. THEOR. THE equal to one another; and if the equal fides be produced, the angles upon the other side of the base shall be equal. Let ABC be an Isosceles triangle, of which the fide AB is equal . a. 3. I. Book 1. to AC, and let the straight lines AB, AC be produced to D and m 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 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 tri А. B C E whole AG, of which the parts AB, AC 4; 3. Ax. are equal; and the remainder BF shall be equal to the remainderCG. 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 BC 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 FBCis equal totheangle 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 it has also been proved that the angle FBCis equal to the 'angle GCB, which are the angles upon the other fide of the base. Therefore the angles at the base, &c. Q. E. D. COROLLARY. Henceevery equilateral triangleis also equiangular. IF the sides also which subtend, or are opposite to, the equal angles shall be equal to one another. |