NOTE 12. Euclid's Proof of I. 6. If two angles of a triangle be equal to one another, the sides also, which subtend the equal angles, shall be equal to one another. B In A ABC let ▲ ACB=L ABC. Then must AB=AC. For if not, AB is either greater or less than AQ From AB cut off BD=AC, and join DC. Then in As DBC, ACB, ::DB=AC, and BC is common, and ▲ DBC= 2 ACB, .. ▲ DBC= ▲ ACB; that is, the less-the greater; which is absurd. .. AB is not greater than AC. Similarly it may be shewn that AB is not less than AC; I. 4. .. AB=AC. Q. E. D. NOTE 13. Euclid's Proof of I. 7. 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 their sides which are terminated in the other extremity of the base equal also. If it be possible, on the same base AB, and on the same side of it, let there be two as ACB, ADB, such that AC=AD, and also BC=BD. Join CD. First, when the vertex of each of the AS is outside the other ▲ (Fig. 1.) ; :: AD=AC, .. 4 ACD= L ADC. But ACD is greater than ▲ BCD ; .. ▲ ADC is greater than ▲ BCD ; much more is ▲ BDC greater than ▲ BCD. I. 5. that is, ▲ BDC is both equal to and greater than ▲ BCD; which is absurd. Secondly, when the vertex D of one of the As falls within the other ▲ (Fig. 2); Then Produce AC and AD to E and F AC=AD. .. L ECDL FDC. I. 5. But ECD is greater than ▲ BCD ; .. FDC is greater than 4 BCD ; much more is 4 BDC greater than 4 BCD. that is, BDC is both equal to and greater than ▲ BCD; which is absurd. Lastly, when the vertex D of one of the As falls on a side BC of the other, it is plain that BC and BD cannot be equal. Q. E. D. NOTE 14. Euclid's Proof of I. 8. 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 must be equal to the angle contained by the two sides of the other. Let the sides of the As ABC, DEF be equal, each to each, that is, AB-DE, AC=DF and BC=EF. Apply the Then Then must ▲ BAC LEDF. ABC to the A DEF. so that pt. B is on pt. E, and BC on EF. :: BC=EF, .. C will coincide with F, and BC will coincide with EF. Then AB and AC must coincide with DE and DF. For if AB and AC have a different position, as GE, GF, then upon the same base and upon the same side of it there can be two As, which have their sides which are terminated in one extremity of the base equal, and their sides which are terminated in the other extremity of the base also equal: which is impossible. .. since base BC coincides with base EF, AB must coincide with DE, and AC with DF; .. BAC coincides with and is equal to I. 7. EDF. Q. E. D. NOTE 15. Another Proof of I. 24. In the As ABC, DEF, let AB=DE and AC=DF, and let BAC be greater than ▲ EDF. Then must BC be greater than EF. Apply the A DEF to the AABC Then BO, OF together are greater than BF, and OC, AO. .. BC, AF. S.E. and AF-AC, I. 20. AC; I. 20. .BF, AC together, .. BC is greater than BF; and.. EF is less than BC. Q. E. D. NOTE 16. Euclid's Proof of I. 26. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz., either the sides adjacent to the equal angles, or the sides opposite to equal angles in each; then shall the other sides be equal, each to each; and also the third angle of the one to the third angle of the other. Let ABC = DEF, and 4 ACB = LDFE; and first, Let the sides adjacent to the equals in each be equal, that is, let BC=EF. Then must AB=DE, and AC=DF, and 4 BAC = LEDF. For if AB be not=DE, one of them must be the greater. Let AB be the greater, and make GB=DE, and join GC. Then in As GBC, DEF, • GB=DE, and BC=EF, and ≤ GBC = ▲ DEF, But .. 4 GCB=LDFE. ▲ ACB= L DFE by hypothesis; .. ▲ GCB= 4 ACB; that is, the less the greater, which is impossible. .. AB is not greater than DE. I. 4. In the same way it may be shewn that AB is not less than DE; .. AB=DE. Then in As ABC, DEF, ·.' AB=DE, and BC=EF, and ▲ ABC= ▲ DEF, I. 4. |