« PreviousContinue »
is neceffary to consider a solid, that is, a magnitude which
has length, breadth, and thickness, in order to understand aright the definitions of a point, line, and superficies ; for these all arise from a solid, and exist in it: The boundary, or boundaries which contain a solid are called superficies, or the boundary which is common to two folids wbich are contiguous, or which divides one folid into two contiguous parts, is called a fuperficies : Thus, if BCGF be one of the boundaries which contain thę solid ABCDEFGH, or which is the common boundary of this solid, and the folid BKLCFNMG, and is there. fore in the one as well as the other solid, is called a superficies, and has no thickness : For if it have any, this thickness mult either be a part of the thickness
H G of the solid AG, or the solid
M BM, or a part of the thickness of
ΕΛΝ cach of them. It cannot be - E part of the thickness of the solid BM; because, if this folid be re.
D moved from the solid AG, the
L superficies BCGF, the boundary of the folid AG, remains still the fame as it was. Nor can it be a
A B K part of the thickness of the folid AG; because, if this be removed from the folid BM, the superficies BCGF, the boundary of the folid BM, does nevertheless remain ; therefore the super, ficies BCGF has no thickness, but only length and breadth.
The boundary of a superficies is called a line, or a line is the common boundary of two fuperficies that are contiguous, or which divides one fuperficies into two contiguous parts : Thus, if BC be one of the boundaries which contain the superficies ABCD, or which is the common boundary of this superficies, and of the superficies KBCL which is contiguous to it, this boundary BC is called a line, and has no breadth : For, if it have any, this must be part either of the breadth of the superficies ABCD, or of the superficies KBCL, or part of cach of them. It is not part of the breadth of the superficies KBCL ; zor, if this superficies be removed from the superbicics ABCD,
the line BC which is the boundary of the superficies ABCD remains the same as it was : Nor can the breadth that BC is supposed to have, be a part of the breadth of the superficies ABCD; because, if this be removed from the superficies KBCL, the line BC which is the boundary of the superficies KBCL does nevertheless remain : Therefore the line BC has no breadth : And be. cause the line BC is in a superficies, and that a superficies has no thickness, as was shewn; therefore a line has neither breadth nor thickness, but only length.
The boundary of a line is called a point, or a point is the common boundary or extremity
B K of the length of KB; for, if the line KB be removed from AB, the point B which is the extremity of the line AB remains the same as it was: Nor is it part of the lengih of the line AB ; for, if AB be removed from the line KB, the point B, which is the extremity of the line KB, does neverthelets remain : Therefore the point B has no length: And becouse a point is in a line, and a line has neither breadth nor thickness, therefore a point has no length, breadth, nor thicknefs. And in this manner the definitions of a point, line, and superficies are to be understood.
DEF. VII. B. I. Instead of this definition as it is in the Greek copies, a more distinct one is given from a property of a plane superficies, which is manifestly supposed in the elements, viz. that a straight line drawn from any point in a plane to any other in it, is wholly in that plane.
DEF. VIII. B. I. It seems that he who made this definition designed that it Mould comprehend not only a plane angle contained by two straight lines, but likewise the angle which fome conceive to be made by a straight line and a curve, or by two curve lines, which meet one another in a plane : But, thoi the meaning of
gles ACD, ACG are equal to one another, which is impossible. Book I. Therefore BD is equal to AC; and by this Propofition BDC is a right angle.
PROP. 3. 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 angles, the angle a 15.1: GHE is therefore equal to the angle AFE which is a right b 4. 1. angle: Draw GK perpendicular to AB; and because the straight lines FK, HG are at right an. A
K B M gles to FH, and
NI KG at right angles to FK; KGO is equal to FH, D
H by Cor. Pr. 2. P 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 cake AP equal to LM, that is, to the double of KG, or AO; and let this be done will the straight line taken be greater than AD: Let this straight line fo taken be AP, and because AP is equal to LM, theretore LM is greater than AD. Which was to be done.
Book I. Bisect AC in F, and draw FG perpendicular to AB; take
CH in the straight line CD equal to AG, and on the contrary fide of AC to that on which AG is, and join FH : Therefore, in the triangles AFG, CFH the fides FA, AG are equal to
FC, CH, each to each, and the angle a 15.1. FAG, that a is, EAB is equal to the
E b 4. 1. angie FCH; wherefore the angle AGF is equal to CHF, and AFG to
two last are equal together to two C H D C 13. 1. right angles", therefore allo AFG, d 14. 1. AFH are equal to two right 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.
PRO P. 5. If two straight lines AB, CD be cut by a third ACE so as to make the interior angles BAC, ACD, on the fame lide of it, together less than two right angles; AB and CD being produced shall meet one another cowards the parts on which are the two angies which are less than two right angles.
At the point C in the straight line CE make a 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 argles
E ECF, FCA are together eb 13. 1. qualb to trvo right angles.
MC F K the angles EAB, FCA are equal to two righe angles. But, by the hypothesis, the
D angles EAB, ACD are to.
L gether less than two right angles; therefore the angle A OG B
H FCA's 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 ihall be greater than the straight line CG: Let
this point be H, and draw HK perpendicular to CF meeting Book I. AB'in L: And because AB, CF contain equal angles with AC on the same fide 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 than CĞ, and therefore is greater than KL, and consequently the point H is in KL produced. Wherefore the straight lines CDH drawn betwixt the points C, H which are on contrary sides of AL, muft neceffarily cut the straight line AB
PRO P. 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 Propofition 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 Me 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 case in the 34. Prop. B. 1. it was thought bet. ter to make use only of the 4. of B. 1.
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.
PRO P. XII. B. II.
N this Proposition only acute angled triangles are mention. Book II.
ed, whereas it holds true of every triangle : And the de. monstrations of the cases omitted are added; Commandine and Clavius have likewise given their demonftrations of these cases.
PRO P. XIV. B. II. In the demonstration of this, some Greek editor has işe norantly inserted the words, “ but if not, one of the two BE,