PROPOSITION I. PROBLEM. TO describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line; it is required to describe an equilateral triangle upon it. From the centre A, at the distance AB, describe a the circle BCD, and from the centre B, at the distance BA, describe the circle ACE; and from the point C, in which the circles cut one another, draw the straight lines b CA, CB to the points A, B; ABC shall be an equilateral triangle. D C Book I. a 3. Postulate. A B E b 1. Post. Because the point A is the centre of the circle BCD, AC is equal to AB; and because the point B is the centre of the c 15. Decircle ACE, BC is equal to BA: but it has been proved that CA finition. is equal to AB; therefore CA, CB are each of them equal to AB; but things which are equal to the same are equal to one anotherd; therefore CA is equal to CB; wherefore CA, AB, BC d 1st Axare 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. PROP. II. PROB. FROM a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC. From the point A to B draw a the straight line AB; and upon it describe the equilateral triangle DAB, and produce the straight lines DA, DB to E and F; from the centre B, at the distance BC, described the circle CGH, and from the centre D, at the distance DG, describe the circle GKL. AL shall be equal to BC. C K A iom. a 1. Post. b 1. 1. H c 2. Post. d 3. Post. F Book I. Because the point B is the centre of the circle CGH, BC is equale to BG; and because D is the centre of the circle GKL, e 15. Def. DL is equal to DG, and DA, DB, parts of them, are equal; f3. Ax. therefore the remainder AL is equal to the remainder f 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. a 2. 1. PROP. III. PROB. FROM the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C, the less. From the point A draw a the straight line AD equal to C; and from the centre A, and at the dis b 3. Post. tance AD, describe b the circle DEF; and because A is the cen c 1. Ax. A F 2 B tre 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 line AE is equal to C, and from AB, the greater of two straight lines, a part AE has been cut off equal to Ċ 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 which the equal sides 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. 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 coinciding 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 impossible. Therefore a 10. Ax the base BC shall coincide with the base EF, and be equal to it. Wherefore the whole triangle ABC shall 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 angles at the base of an isosceles triangle are equal to one another; and, if the equal sides be produced, the angles upon the other side of the base shall be equal. Let ABC be an isosceles triangle, of which the side AB is Book 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. a 3. 1. b 4. 1. In BD take any point F, and from AE the greater, cut off AG equala to AF, the less, and join FC, GB. F D each to each; and A B C G E Because AF is equal to AG, and AB to AC, the two sides FA, AC are equal to the two GA, AB, they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is equal to the base GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are equal to the remaining angles of the other, each to each, to which the equal sides are opposite; viz. the angle ACF to the angle ABG, and 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 c 3. Ax. remainder 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 BC is common to the two triangles BFC, CGB; wherefore the triangles are equal b, 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 subtend, or are opposite to, the equal angles shall be equal to one another. Let ABC be a triangle having the angle ABC equal to the Book I.. 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 D A the other; let AB be the greater, and from it cut a off DB a 3. 1. 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 DB, BC are equal to to the two AC, CB, each to each; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle ACB, the less to the greater; which is absurd. Therefore AB is not unequal to AC, that is, it is equal to Wherefore, if two angles, &c. Q. E. D. it. B COR. Hence every equiangular triangle is also equilateral. b 4. 1. C PROP. VII. THEOR. UPON the same base, and on the same side of it, See N. 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 equal a to the angle ADC: but the angle ACD is greater than the angle BCD; therefore the angle ADC is greater also than BCD; A C D B much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal a to a 5. 1. the angle BCD; but it has been demonstrated to be greater than it, which is impossible. |