Book I. Araight line CFD is nearer to the straight line AB at the point F
Nthan at the point C, that is CF comes

nearer to AB from the point C to F.
but because DB is greater than FE, the

Araight line CFD is further from AB
at the point Dthan at I, that is FD goes

further from AB from F to D. there-
fore the st: aight line CFD first comes

nearer to the straight line AB, and then
froes further from it, before it cuts it, which is impoflible, and the
same thing will follow, if FE be said to be greater than CA, or DB.
therefore FE is not uncqual to AC, that is, it is equal to it.

PRO P. 2. If two equil straight lines AC, BD be each at right angles to te fame straight line AB; the straight line CD which joins their extremities makes right angles with AC and BD.

Join AD, BC; and because in the triangles CAB, DBA, CA,

AB are equal to DB, BA, and the angle CAB equal to the angle 2. 4. 1.

DBA; the base BC is equal to the base AD, and in the triangles ACD, BDC, AC, CD are equal to BD, DC, and the bafe AD is equal to the base EC, therefore the an

F D b. 8. 1. gle ACD is equal to the angle BDC.

C from any poinç E in AB draw EF unto

G CD, at right angles to AB; therefore, by Prop. 1. EF is equal to AC, or BD; wherefore, as has been just now thewn, A E B the angle ACF is equal to the angle EFC. in the same manner the angle BDF is equal to the angle

EFD; but the angles ACD, BDC are equal, therefore the angles Lic. Def... EPC and LFD are equal, and right angles; wherefore also the

angles ACD, BDC are right angles.

Cor. Hence, if two straight lines AB, CD be at right angles to the same straight line AC, and if betwixt them a straight line ED be drawn at right angles to either of them, as to AB; then BD is equal to AC, and BDC is a right angle.

IF AC be not equal to BD, take BG equal to AC, and join CG. therefore, by this Pros a'ition, the angle ACG is a right angle; but ACD is also a right angle, wherefore the angles ACD, ACG are

equal to one another, which is impoflible. therefore BD is equal Book I. to AC; and by this Proposition BDC is a right angle.

PRO P. 3


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If two straight lines which contain an angle be produced, there may be found in either of them a point from which the perpendicular drawn to the other shall be greater than any given straight line.

Let AB, AC be two straight lines which make an angle with one another, and let AD be the given straight line ; a point may be found either in AB or AC, as in AC, from which the perpendicular drawn to the other AB shall be greater than AD.

In AC take any point E, and draw EF perpendicular to AB; produce AE to G so that EG be equal to AE; and produce FE to H, and make EH equal to FE, and join HG. because, in the triangles AEF, GEH, AE, EF are equal to GE, EH, each to each, and contain equal a angles, the angle GHE is therefore equal to the angle AFE which is a right angle. draw GK perpendicular 6.4.1, to AB; and because

А. F K B M the straight lines FK, HG are at right an-N

E gles to FH, and KG 0! at right angles to D!

H FK; KG is equal to pl FH, by Cor. Pr. 2. that is to the double of FE. in the same manner, if AG be produced to L so that GL be equal to AG, and LM be drawn perpendicular to AB, then LM is double of GK, and so on. In AD take AN equal to Fe, and AO equal to KG, that is to the double of FE, or AN; also take AP equal to LM, that is to the double of KG, or AO; and let this be done till the straight line taken be greater than AD. let this straight line so taken be AP, and because AP is equal to LM, therefore LM is greater than AD. Which was to be done.


PRO P. 4. If two straight lines AB, CD make equal angles EAB, ECD with another straight line EAC towards the fame parts of it; AB and CD are at right angles to some straight line.

Bisect AC in F, and draw FG perpendicular to AB; take CH in the straight line CD equal to AG and on the contrary side of AC 10

2. 15. 1.

b. 4. 1.

c. 13. I.

d. 14. 1.

Book I. thaton which AG is, and join FH. therefore, in the triangles AFG,

CFH the sides FA, AG are equal to FC, CH, each to each, and
the angle FAG, that is EAB is equal to

the angle FCH ; wherefore b the angle
AGF is equal to CHF, and AFG to the
angleCFH. to these last add the common

GA / B.
angle AFH, therefore the two angles
AFG, AFH are equal to the two angles
CFH, HFA which two last are equal to-
gether to two right angles. ; therefore CH
alfo AFG, AFH are equal to two right

D angles, and consequently GF and FH are in one straight line. and because AGF is a right angle, CHF which is equal to it is also a right angle. therefore the straight lines AB, CD are at right angles to GH,

PROP. 5. If two straight lines AB, CD be cut by a third ACE so as to make the interior angles BAC, ACD, on the same side of it, together less than two right angles; AB and CD being produced thall meet one another towards the parts on which are the two angles which are less than two right angles.

At the point C in the straight line CE make the angle ECF equal to the angle EAB, and draw to AB the straight line CG at right angles to CF. then because the angles ECF, EAB are equal to one another, and that the angles ECF, FCA are toge

E b. 13, 1, ther equal b tv two right angles, the angles EAB, FCA


F K are equal to two right angles. but, by the hypothesis, the N

D angles EAB, ACD are toge

L ther less than two right angles, therefore the angle FCA AOG


H is greater than ACD, and CD falls between CF and AB. and because CF and CD make an angle with one another, by Prop. 3. a point may be found in either of thein CD from which the perpendicular drawn to CF shall be greater than the straight line CG. let this point be H, and draw HK ferpendicular to CF meeting AB in L. and becaufe AB, CF contain equal angles with AC on the same side of it, by Prop. 4. AB and

2. 2 3. 1.

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CF are at right angles to the straight line MNO which bisects AC Book I, in N and is perpendicular to CF. therefore, by Cor. Prop. 2. CG and KL which are at right angles to CF are equal to one another. and HK is greater than CG, and therefore is greater than KL, and consequently the point H is in KL produced. Wherefore the straight line CDH drawn betwixt the points C, H which are on contrary sides of AL, muft neceffarily cut the straight line AB.

PROP. XXXV. B. I. The Demonstration of this Proposition is changed, because if the method which is used in it was followed, there would be three cases to be separately demonstrated, as is done in the translation from the Arabic; for in the Elements no case of a Proposition that requires a different demonstration ought to be omitted. On this account we have chosen the method which Monf. Clairault has given, the first of any, as far as I know, in his Elements, page 21. and which afterwards Mr. Simpson gives in his, page 32. but whereas Mr. Simpson makes use of Prop. 26. B. 1. from which the equality of the two triangles does not immediately follow, because to prove that, the 4. of B. 1. must likewise be made use of, as may be seen, in the very fame cafe, in the 34. Prop. B. 1. it was thought better to make use only of the 4. of B. I.

PRO P. XLV. B. I. The straight line KM is proved to be parallel to FL from the 33. Prop. whereas KH is parallel to FG by construction, and KHM, FGL have been demonstrated to be straight lines. a Corollary is added from Commandine, as being often used.


B. II.

Book II.

N this Proposition only acute angled triangles are mentioned,

whereas it holds true of every triangle. and the Demonítrations of the cases omitted, are added; Commandine and Clavius have likewise given their Demonstrations of these cases.

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PROP. XIV. B. II. In the Demonstration of this, fome Greek Editor has ignorantly inserted the words, “ but if not, one of the two BE, ED is the

Book II. “ greater ; let BE be the greçter and produce it to T," as if it was of Wany consequence whether the greater or lesser be produced. there

fore instead of these words, there ouzht to be read only, “ but if

not, produce BE to F.”

Book III.



EVERAL Authors, especially among the modern Mathemati

cians and Logicians, inveigh too severely against indirect, or Apagogic Demonstrations, and sometimes ignorantly enough ; not being aware that there are some things that cannot be demonstrated any other way. of this the present Proposition is a very clear instance, as no direct demonstration can be given of it. because, befides the Definition of a circle, there is no principle or property relating to a circle antecedent to this Problem, from which cither a direct or inchirect Demonstration can be deduced. wherefore it is necessary that the point found by the construction of the Problem be proved to be the center of the circle, by the help of this Definition, and some of the preceding propofitions, and because in the Demonstration, this Proposition inust be brought in, viz. straight lines from the center of a circle to the circumference are equal, and that the point found by the construction cannot be assumed as the center, for this is the thing to be demonstrated; it is manifest some other point must be affumed as the center; and if from this assumption an absurdity follows, as Euclid demonstrates there must; then it is not true that the point assumed is the center; and as any point whatever was assumed, it follows that no point, except that found by the construction can be the center. from which the necessity of an indirect Demonstration in this case is evident.


XIII. B. III. As it is much easier to imagine that two circles may touch one another within in more points than one, upon the same side, than upon opposite sides; the figure of that case ought not to have been omitted ; but the construction in the Greek text would not have suited with this figure so well, because the centers of the circles must have been placed near to the circumferences. on which account another construction and demonstration is given which is the fame with the second part of that which Campanus has translated

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