« ForrigeFortsett »
*INTEnc.12. Prolong* AM to an unlimited length on the side of Z. At the
o, 3 point A in the straight line AM describet the rectilinear figure
ACOM, equal in all respects to ABNM, and having its base AC (at which are the angles that are equal to BAM, ABN), adjacent to the base AB. In like manner at the points B and C, make the rectilinear figures BP, and CQ, equal in all respects to ABNM or ACOM; and so on continually on both sides, for as many as shall at any time be desired.
† Constr. Because the angles MAB, NBA, MAC, OCA, &c. aref equal to one another and each less than a right angle, and the straight lines AB, AC, BD, CE, &c. are equal to one another; these equal straight lines form a series as in Prop. XXVIII B; of 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 it may cut, an interior angle each way, greater" than a given angle 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 was either equal tof or greater thanf half BAC, the straight lines bisecting the angles of the series must have met in the axis, 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 one pair of angular points to another, have* failed to make an interior angle on the side towards Z, greater than BAZ; neither can it havet re-met the axis. Wherefore (because WX can in no 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 MX; and the angle MRB must bef equal to AMR and consequently a right angle, and BR equal to AM ; and because 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 falls on the extremity of the base B ist equal to AM, and BR 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
* I. 28 B.
# I. 28 C.
* I. 28 C.
# I. 28 C. Cor. 3.
f I. 28 A.
* I. 12.
* I. 16.
+ INTERC. l. Cor. 2. f I. 28 A. * I. 19.
* I. 25.
WX; wherefore, because WX cuts the series, some or other of these sides must meet its base, as FH in U; which is impossible, because every one of these sides is" parallel to its base. The assumptiont, 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,
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.
CE equal to AB; and join BD, BE. Because
PROPOSITION XXVIII E.
THEOREM-In every right-angled triangle the two acute angles are together not less than a right angle.
Let DAB be a triangle, of which the angle 2 C *1-17 Cor.3. DAB is a right angle. The two" acute angles ADB, DBA, are together not less than a right A B angle. # I. 10. From the point B drawit a straight line BC at right angles to 1. I. 3. BA; and makes BC equal to AD. And join CD. * Constr. Because DABC is" a tessera having the angles at the base AB
# I. 28 D. right angles, the side CD isł 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 fl. 24. angle CBD ; for if it was less, the third side AB must bef 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 *INTEnc. l. 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 C ti.17.Cor.3, two: acute angles, let AB be a side of the o N triangle which lies between two such. From * I. 23. the opposite angular point draw” a straight A D B line CD perpendicular to AB; and because the side AB lies beti.23.Cor.3. tween two acute angles, CD willf 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 H
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 A. E ** 12 ACB and ACD are" together equal to two right angles. And the angles ACB, ABC, BAC are (by Cor. 1 above) together not less than two right angles. Whence ACB, 5 C D ABC, BAC, are together not less than ACB and ACD; for if *INTERC, 1 they were less, they must also bet 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 bef less than ACB and ACD, and they are not.
PROPOSITION xxviii F.
THEOREM.–If two straight lines nhich are in the same plane are
plane, be met by the straight line AC; and let the two interior A o
ETC G H I K. D 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. *INTERc12. Prolong* DC to E; and at A maket the angle CAF equal to §§2. ACE. In CD take; CG equal to CA; and join GA. t I. 3. Because the angle CAF is equal to ACE, the angles CAF and Jono. 1. ACD are” together equal to ACE and ACD; but ACE and
.4. §. ACD aret together equal to two right angles; wherefore CAF