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PROP. XI. B. I. A Corollary is added to this Proposition, which is neceffary to Prop. 1. B. 11. and otherwise.
PROP. XX. and XXI. B. I. Proclus in his Commentary relates that the Epicureans derided this Proposition, as being manifest even to asses, and needing no Demonftration; and his answer is, that tho' the truth of it be manifest to our senses, yet it is science which must give the reason why two sides of a triangle are greater than the third. but the right answer to this objection against this and the 21. and some other plain Propositions, is, that the number of Axioms ought not to be encreased without necessity, as it must be if these Propositions be not demonstrated. Mons. Clairault in the Preface to his Elements of Geometry published in French at Paris Ann. 1741. says that Euclid has been at the pains to prove that the two sides of a triangle which is included within another are together less than the two fides of the triangle which includes it; but he has forgot to add this condition, viz. that the triangles must be upon the fame base; because unless this be added, the sides of the included triangle may be greater than the sides of the triangle which includes it, in any ratio which is less than that of two to one. as Pappus Alexandrinus has demonstrated in Prop. 3. B. 3. of his Mathematical Collections.
Some Authors blame Euclid because he does not demonstrate that the two circles made use of in the construction of this Problem must cut one another. but this is very plain from the determination he has given, viz. that any two of the straight lines DF, FG, GH must be greater than the third. for who is so dull, tho' only beginning to learn the Elements, as not to perceive that the circle described from the center F, at the distance FD, must meet FH betwixt F and H,DM FG H because FD is less than FH; and that, for the like reason, the circle described from the center G, at the distance GH or GM muft meet DG betwixt D and G; and
Book I. that these circles must meet one another, because FD and GH are
together greater than FG? and this
PROP. XXIV. B. I.
To this is added “ of the two fides DE, DF, let DE be that & which is not greater than the other;" that is, take that fide of the two DE, DF which is not greater than the other, in order to make with it the angle EDG equal to
Mr. Thomas Simpson in p. 262. of
PROP. XXIX. B. I.
not to be properly placed among the Axioms, as, indeed, it is not Book I. self-evident; but it may be demonstrated thus.
DEFINITION The distance of a point from a straight line, is the perpendicular drawn to it from the point.
DEF. One straight line is said to go nearer to, or further from another straight line, when the distance of the points of the first from the other straight line become less or greater than they were; and two straight lines are said to keep the fame distance from one another, when the distance of the points of one of them from the other is always the same.
AXIO M. A straight line cannot first come nearer to another straight line, and then go further from it, before it cuts it; and, in like manner, a
B straight line cannot go further from, another straight line, and then come
nearer to it; nor can a straight line F
H keep the same distance from another straight line, and then come nearer to it, or go further from it; or a Praight line keep always the fame direction.
For example, the straight line ABC cannot first come nearer to the straight line DE, as from the point A to the point B, and then, from the A.
See the forma
C gure above. point B to the point C, go further D
E from the fame DE. and, in like man F G ner, the straight line FGH cannot go
H further from DE, as from F to G, and then, from G to H, come nearer to the fame DE. and so in the last case as in fig. 2.
PROP. If two equal straight lines AC, BD be each at right angles to the same straight line AB; if the points C, D be joined by the straight line CD, the straight line EF drawn from any point Ein AB unto CD, at right angles to AB, shall be equal to AC, or BD.
If EF be not equal to AC, one of them must be greater than the other; let AC be the greater; then because FE is less than CA, the
Book 1. straight line CFD is nearer to the straight line AB at the point F than at the point C, that is CF comes
A E B
a. 4. I.
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 to the angle DBA; the base BC is equal a 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 anb. 8. 1. gle ACD is equal o to the angle BDC.
F D 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 shewn, 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 C.10 Def.I. EFC and EFD 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 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 impoflible. therefore BD is equal Book I. 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 a angles, the angle GHE is therefore equal 6 4. 15. 9. to the angle AFE which is a right angle. draw GK perpendicular to AB; and because the straight lines FK, A F K B M HG are at right an-NI
3 gles to FH, and KG OI at right angles to Del
PROP. 4. 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.
· 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