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Book I. straight line AB at the point F than 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 C straight line CFD is further from AB at the point D than at F, that is, FD goes further from AB from F to D: Therefore the straight line CFD first comes nearer to the straight line AB, and then goes further from it, before it cuts it: which is impossible: If FE be said to be greater than CA, or DB, the straight line CFD first goes further from the straight line AB, and then comes nearer to it, which is also impossible. Therefore FE is not unequal to AC, that is, it is equal to it.



IF two equal straight lines AC, BD be each at right angles to the same 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 4. 1. to the angle 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 base AD is equal to

the base BC. Therefore the an8. 1. gle ACD is equal to the angle BDC: From any point E in AB draw EF unto CD, at right angles to AB; therefore by Prop. 1. EF


is equal to AC, or BD; wherefore, as has been just now shown, 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 • 10 Def. 1. EFC and EFD are equal, and right angles; wherefore also the angles ACD, BDC are right angles.

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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 BD 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 proposition, 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 impossible. Boox I.
Therefore BD is equal to AC; and by this proposition
BDC is a right angle.


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.

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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 angles, 15. 1. the angle GHE is therefore equal to the angle AFE which is a right angle: Draw GK perpendicular to AB; and because the straight lines FK, HG are at right angles to FH, and

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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


IF two straight lines AB, CD make equal angles EAB ECD with another straight line EAC towards the same parts of it: AB and CD are at right angles to some straight line.


4. 1.



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 to that on 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, 15. 1. thata is EAB, is equal to the angle

b 4. 1. FCH; wherefore the angle AGF

is equal to CHF, and AFG to the angle CFH: To these last add the common angle AFH; therefore the two angles AFG, AFH are equal to the two angles CFH, HFA, which two last are equal 13. 1. together to two right anglese:

therefore also AFG, AFH are equal






14. 1. to two right angles, and consequentlyd 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.


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, shall meet one another towards the parts on which are the two angles, which are less than two right angles.

* 23. 1. At the point C, in the straight line CE, makea 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

13. 1. ther equal to two

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right angles; there

fore the angle FCA 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 them CD, from which the perpendicular

drawn to CF shall be greater than the straight line CG. Let this point be H, and draw HK perpendicular to CF, meeting AB in L:And because AB, CF contain equal angles with AC on the same side of it, by Prop. 4. AB and CF are at right angles to the straight line MNO, which bisects AC 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 thart 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, must necessarily cut the straight line AB.


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 Mons. 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. I. from which the equality of the two triangles does not immediately follow, because, to prove that, the 4th of B. I. must likewise be made use of, as may be seen in the very same case in the 34th Prop. B. I. it was thought better to make use only of the 4th of B. I.


THE straight line KM is proved to be parallel to FL from the 33d 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.


In this proposition only acute angled triangles are mentioned, whereas it holds true of every triangle; and the demonstrations of the cases omitted are added: Commandine and Clavius have likewise given their demonstrations of these cases.

Book I.



Book II.



IN the demonstration of this, some Greek editor has ignorantly inserted the words, " but if not, one of the two "BE, ED, is the greater: Let BE be the greater, and produce it to F:" as if it was of any consequence whether the greater or lesser be produced: Therefore, instead of these words, there ought to be read only, "but if not, "produce BE to F."


Book III. SEVERAL authors, especially among the modern mathematicians and logicians, inveigh too severely against indirect or apogogic 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, besides the definition of a circle, there is no principle or property relating to a circle antecedent to this problem, from which either a direct or indirect demonstration can be deduced: Wherefore it is necessary that the point found by the construction of the problem be proved to be the centre of the circle, by the help of this definition, and some of the preceding propositions: And because, in the demonstration, this proposition must be brought in, viz. straight lines from the centre of a circle to the circumference are equal, and that the point found by the construction cannot be assumed as the centre, for this is the thing to be demonstrated; it is manifest some other point must be assumed as the centre: and if from this assumption an absurdity follows, as Euclid demonstrates there must, then it is not true that the point assumed is the centre; and as any point whatever was assumed, it follows that no point, except that found by the construction, can be the centre, from which the necessity of an indirect demonstration in this case is evident.


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 centres of the circles must have been placed near to the circumferences; on which

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