*Interc.12. Prolong* AM to an unlimited length on the side of 2. At the Cor. 6. point A in the straight line AM describet the rectilinear figure +1.22.Cor.3. ACOM, equal in all respects to ABNM, and having its base AC Because the angles MAB, NBA, MAC, OCA, &c. are equal which series AZ is the axis. Let then a straight line WX of unlimited length both ways, be made to move along the axis, from the vertex A towards Z, keeping ever at right angles to the axis ; till it cuts the axis in M. But it has been shown that on the side towards A, it will at all times during such motion (after passing through BC) make with such parts of the series as * 1. 28 B. it may cut, an interior angle each way, greater* than a given angle Cor. 4. ABC. And on the side towards Z, it will at all times make an interior angle each way, greater than a given angle BAZ. For when WX arrives at any of the angular points (as D, E), the next of the equal straight lines of the series (as DF) cannot fail to make with WX an angle EDF greater than BAZ which is half BAC, unless the angle BDE at the cusp be equal to or greater than half BAC; which last cannot be, because if BDE the angle at the cusp + I. 28 C. was either equal tot or greater than# half BAC, the straight Cor. 1. lines bisecting the angles of the series must have met in the axis, I 1. 28 C. in the part between the point where WX in passing through D and E cuts the axis, and the vertex; and they have not. And for a like reason, it cannot at any time during its passing from * I. 28 C. one pair of angular points to another, have* failed to make an Cor. 2. interior angle on the side towards Z, greater than BAZ; neither + I. 28 C. can it havet re-met the axis. Wherefore (because WX can in no Cor. 3. way have previously ceased to cut the series and make with it on both sides of WX an interior angle greater than a given angle) it will cut the series when it arrives at the point M; as for instance in U and V. Whence the angles opposite to the base of the tessera ABNM cannot be right angles. For if this be disputed, let it be assumed that they are right angles. But if so, then in the tessera whose base is AB, because AMX is a right angle, the side opposite to the base must be in the same straight line with 11. 28 A. and the angle MRB must be f equal to AMR and con sequently a right angle, and BR equal to AM; and because * I. 12. MRB, BRX are* together equal to the sum of two right angles and MRB is a right angle, BRX must be a right angle ; and because in the tessera whose base is BD, the side which + Constr. falls on the extremity of the base B ist equal to AM, and BR Interc. 1. is equal to AM, such side must bef equal to BR and coincide with it; and for like reasons the sides opposite to the bases BD, DF, &c. of the tesseras which follow in order, must coincide with further portions RS, ST, &c. of the straight line MX; WX; wherefore, because WX cuts the series, some or other of these sides must meet its base, as FH in U; which is impossible, * I. 28 A. because every one of these sides is* parallel to its base. The as+1.Nom.26. sumptiont, therefore, which involves this impossible consequence, cannot be true; or the angles opposite to the base of the tessera ABNM, cannot be right angles. And by parity of reasoning, the like may be proved of every other tessera where the angles at the base are less than right angles. Wherefore, universally, if the angles at the base of a tessera be less than right angles, &c. Which was to be demonstrated. Cor. 1. If the angles at the base of a tessera be right angles, the angles opposite to the base cannot be less than right angles. For if they were, then (by considering the opposite side as a base) there would be a tessera having the angles at the base less than right angles, and the angles opposite to the base right angles; which (by the Prop. above) is impossible. i 1.3. Cor. 2. If the angles at the base AB of a tessera be right angles, the side opposite to the base is not greater than the base. For if CD be greater than AB, from CD cut off# DE DEB is the exterior angle of the triangle ECB • I. 16. it is* greater than the interior and opposite angle A B + INTERC. 1. C; and consequently greatert than the angle EDA which ist Cor. 2. equal to C, and still more than the angle EDB which is less than | 1, 28 A. I. 19. EDA; wherefore BE the side opposite to EDB is* less than BD the side opposite to DEB. But if CE is equal to AB, then + Hyp. because AD ist equal to CB, and AB to CE, the triangles BAD, ECB must have the sides AD, AB equal to the sides CB, CE *Hyp. respectively. And because the angle BAD ist a right angle, and the angle BCE (by Cor. 1 above) is not less than a right angle, the angle BCE is not less than the angle BAD; for if it was less than BAD, it must be less than a right angle, and it is not. Whence the third side BE must be not less than the third side * J. 25. BD; because if it was less, the angle BCE must be* less than the angle BAD, and it is not. Which is impossible, for BE has been shown to be less than BD. It is not true, therefore, that CE, which is a part of CD, is equal to AB; that is, that CD is greater than AB. D с A PROPOSITION XXVIII E. are together not less than a right angle. Let DAB be a triangle, of which the angle •1.17.Cor.3. DAB is a right angle. The two* acute angles B angle. + 1. 10. From the point B drawt a straight line BC at right angles to 11. 3. BA; and maket BC equal to AD. And join CD. • Constr. Because DABC is* a tessera having the angles at the base AB + 1. 28 D. right angles, the side CD ist not greater than AB. Whence in Cor. 2. the triangles DAB, BCD, the sides DA, DB are equal to the sides BC, BD respectively, and the third side AB is not less than the third side CD; wherefore the angle ADB is not less than the $1. 24. angle CBD; for if it was less, the third side AB must be less than the third side CD, and it is not. And because the angle ADB is not less than CBD, add to each the angle DBA, and the sum of ADB and DBA will be not less than the sum of CBD and DBA ; for if it was less, then by taking DBA from each, the *INTERC. 1. remaining angle ADB must be* less than the remaining angle Cor. 8. CBD, and it is not. But the sum of the angles CBD and DBA + Constr. is the angle CBA which ist a right angle; therefore the sum of ADB and DBA is not less than a right angle. And by parity of reasoning, the like may be proved in every other right-angled triangle. Wherefore, universally, in a rightangled triangle the two acute angles are together not less than a right angle. Which was to be demonstrated. Cor. 1. The three angles of every triangle are together not less than two right angles. For because every triangle has at the least 11.17.Cor.3. two acute angles, let AB be a side of the triangle which lies between two such. From D * I. 23. the opposite angular point draw* a straight line CD perpendicular to AB; and because the side AB lies be+1.23.Cor.3. tween two acute angles, CD willt fall between the extremities of AB. And because the triangle ADC is right-angled at D, the angles DAC, ACD (by the Proposition above) are together not less than a right angle. And for the like reason the angles BCD, DBC are together not less than a right angle. Wherefore the B H н E less than two right anglos. Whence ACB, V. angles DAC, ACD, BCD, DBC are together not less than two right angles. But the angles ACD, BCD are together equal to the angle ACB; therefore the angles DAC, ACB, DBC are together not less than two right angles. COR. 2. If one side of a triangle be prolonged, the two interior opposite angles are together not less than the exterior angle. For if BC be prolonged to D, the angles А • I. 12. ACB and ACD are* together equal to two B C ABC, BAC, are together not less than ACB and ACD; for if + INTERC. 1 they were less, they must also beț less than two right angles, and Cor. 2. they are not. And taking away the common angle ACB, the angles ABC and BAC are together not less than ACD; for if they were less, then by adding ACB to each, the angles ACB, INTERC. 1. ABC, BAC must be less than ACB and ACD, and they are not. Cor. 6. PROPOSITION XXVIII F. met by a third, and the two interior angles on the one side of this Let the two straight lines AB and CD which are in the same plane, be met by the straight line AC; and let the two interior D B G K Е С angles CAB, ACD on the one side of AC, be together less than two right angles. AB and CD being continually prolonged. shall at length meet on the side of B and D. *INTERC.12. Prolong* DC to E; and at A maket the angle CAF equal to Cor. 6. +1.22.Cor.2. ACE. In CD taket CG equal to CA; and join GA. 1 1.3. Because the angle CAF is equal to ACE, the angles CAF and *INTERC. 1. ACD are* together equal to ACE and ACD; but ACE and Cor. 4. +1. 12. ACD aret together equal to two right angles; wherefore CAF |