5. 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. (I. 4.) 6. And because the whole AF is equal to the whole AG, of which the parts AB, AC, are equal, (hypoth.) The remainder BF is equal to the remainder CG. 7. And FC was proved to be equal to GB. (ax. 3.) 8. Therefore the two sides BF, FC, are equal to the two sides CG, GB, each to each. 9. And the angle BFC was proved equal to the angle CGB. 10. Therefore the triangles BFC, CGB, are equal; and their other angles are equal, each to each, to which the equal sides are opposite. (I. 4.) 11. Therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. 12. And since it has been demonstrated that the whole angle ABG is equal to the whole angle ACF, and that the parts of these, the angles CBG, BCF, are also equal, 13. Therefore the remaining angle ABC is equal to the remaining angle ACB. (ax. 3.) Which are the angles at the base of the triangle ABC. 14. And it has been proved that the angle FBC is equal to the angle GCB. (demonst. 11.) Which are the angles upon the other side of the base. Conclusion. Therefore, the angles at the base, &c. (see Enunciation.) Which was to be shown. Corollary. Hence every equilateral triangle is also equiangular. PROPOSITION 6.-THEOREM. 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. (References-Prop. I. 3, 4.) Hypothesis. Let ABC be a triangle having the angle ABC equal to the angle ACB. Sequence. The side AB shall be equal to the side AC. (False Hypothesis.)-For if AB be not equal to AC, one of them is greater than the other. Let AB be the greater. Construction.-1. From AB the greater, cut off a part DB, equal to AC the less. (I. 3.) 2. Join DC. Demonstration.-1. Because in the triangles DBC, ACB, DB is assumed to be equal to AC, and BC is common to both, 2. Therefore the two sides DB, BC, are equal to the two sides AC, CB, each to each. 3. And the angle DBC is equal to the angle ACB. (hypoth.) 4. Therefore the base DC is equal to the base AB. (I. 4.) 5. And the triangle DBC is equal to the triangle ACB. (I. 4.) The less to the greater, which is absurd. B A 6. Therefore AB is not unequal to AC, that is, it is equal to it. Conclusion.—Wherefore, if two angles, &c. Q. E. D.* Corollary.-Hence every equiangular triangle is also equi lateral. PROPOSITION 7.-THEOREM. Upon 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. (References-Prop. I. 5; ax. 9. Hypothesis.-1. Let the triangles ACB, ADB, upon the same base AB, and on the same side of it, have, if possible, 2. Their sides CA, DA, terminated in the extremity A of the base, equal to one another. 3. And their sides CB, DB, terminated in the extremity B of the base, likewise equal to one another. Case I.-Let the vertex of each triangle be without the other triangle. Construction.-Join CD. Demonstration.-1. Because AC is A assumed to be equal to AD. (hyp.) D 2. The triangle ADC is an isosceles triangle, and the angle ACD is therefore equal to the angle ADC. (I. 5.) 3. But the angle ACD is greater than the angle BCD. (ax. 9.) 4. Therefore the angle ADC is also greater than BCD. 5. Much more then is the angle BDČ greater than BCD. * Q. E. D. is an abbreviation for quod erat demonstrandum, that is, "which was to be shown or proved." 6. Again, because BC is assumed to be equal to BD, (hyp.) 7. The triangle BCD is an isosceles triangle, and the angle BDC is equal to the angle BCD. (I. 5.) 8. But the angle BDC has been shown to be greater than the angle BCD. (dem. 5.) 9. Therefore the angle BDC is both equal to, and greater than the same angle BCD, which is impossible. Case II.-Let the vertex of one of the triangles fall within the other. Construction.-Produce AC, AD, to E and F, and join CD. Demonstration.-1. Because AC is assumed to be equal to AD. (hyp.) 2. The triangle ADC is an isosceles triangle, and the angles ECD, FDC, upon the other side of its base CD, are equal A to one another. (I. 5.) E B 3. But the angle ECD is greater than the angle BCD. (ax. 9.) 4. Therefore the angle FDC is likewise greater than BCD. 5. Much more then is the angle BDC greater than BCD. 6. Again, because BC is assumed to be equal to BD. (hyp.). 7. The triangle BDC is an isosceles triangle, and the angle BDC is equal to the angle BCD. (I. 5.) 8. But the angle BDC has been shown to be greater than the angle BCD. (dem. 5.) 9. Therefore the angle BDC is both equal to, and greater than the same angle BCD, which is impossible. Conclusion. Therefore, upon the same base, &c. Q. E. D. PROPOSITION 8.-THEOREM. 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. Or, If two triangles have three sides of the one respectively equal to the three sides of the other, they are equal in every respect. (References-Prop. I. 7; ax. 8.) Hypothesis. Let ABC, DEF, be two triangles which have, 1. The two sides AB, AC, equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF. 2. And the base BC equal to the base EF. Sequence. The angle BAC shall be equal to the angle EDF. Demonstration.-1. For if the triangle ABC be applied to DEF, 2. So that the point B may be on E, and the straight line BC on EF, 3. The point C shall A coincide with the point F, because BC is equal to EF. (hyp.) 4. Therefore BC coinciding with EF, BA and AC shall coincide with ED and DF. G 5. For if the base BC coincides with the base EF, 6. But the sides BA, AC, do not coincide with the sides ED, DF, but have a different situation, as EG, GF, 7. Then upon the same base, and on the same side of it, there will be two triangles, which have their sides terminated in one extremity of the base equal to one another, and likewise their sides, which are terminated in the other extremity. But this is impossible. (I. 7.)' 8. Therefore if the base BC coincides with the base EF, the sides BA, AC, must coincide with the sides ED, DF. 9. Therefore the angle BAC coincides with the angle EDF, and is equal to it. (ax. 8.) 10. Also the triangle ABC coincides with the triangle DEF, and is therefore equal to it in every respect. (ax. 8.) Conclusion. Therefore, if two triangles, &c. Q.E.D. PROPOSITION 9.-PROBLEM. To bisect a given rectilineal angle, that is, to divide it into two equal parts. (References-Prop. I. 1, 3, 8.) Given.-Let BAC be the given rectilineal angle. AB. 2. From AC cut off AE equal to AD. (I. 3.) 3. Join DE. 4. Upon DE, on the side remote from A, describe an equilateral triangle DEF. (I. 1.) 5. Join AF. B D Then the straight line AF shall bisect the angle BAC. Proof.-1. Because AD is equal to AE (const.), and AF is common to the two triangles DAF, EAF; 2. The two sides DA, AF, are equal to the two sides EA, AF, each to each; 3. And the base DF is equal to the base EF; (const.) 4. Therefore the angle DAF is equal to the angle EAF. (I.8.) Conclusion. Therefore the given rectilineal angle BÀC is bisected by the straight line AF. Q. E. F. PROPOSITION 10.-PROBLEM. To bisect a given finite straight line, that is, to divide it into two equal parts. (References-Prop. I. 1, 4, 9.) Given. Let AB be the given straight line. the equilateral triangle ABC. (I. 1.) 2. Bisect the angle ACB by the straight line CD. (I. 9.) Then AB shall be cut into two equal parts in the point D. Proof.-1. Because AC is equal to CB (const.), and CD common to the two tri- A angles ACD, BCD; D B 2. The two sides AC, CD, are equal to the two sides BC, CD, each to each; 3. And the angle ACD is equal to the angle BCD; (const.) 4. Therefore the base AD is equal to the base DB. (I. 4.) Conclusion. Therefore the straight line AB is divided into two equal parts in the point D. Q. E. F. PROPOSITION 11.-PROBLEM. To draw a straight line at right angles to a given straight line from a given point in the same. (References-Prop. I. 1, 3, 8; ax. 1; def. 10.) Given.-Let AB be the given straight line, and C a given point in it. Sought. It is required to draw a straight line from the point C at right angles to AB. Construction.-1. Take any point D in AC. 2. Make CE equal to CD. (I. 3.) 3. Upon DE describe the equilateral triangle DFE. (I. Į.) 4. Join FC. Then FC shall be at right angles to AB. |