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way, though produced ever so far, are || to one another (Def. 34); .-. AB is || to CD.

Wherefore, if a st. line, &c.—Q. E. D.

It is clear from the Definition, though Euclid has stated the limitation neither in this nor the following Proposition, that the two lines upon which the other falls should be in the same plane. At the same time, the oversight, if such it be, is so palpable, that it cannot mislead; not to mention that, in the first six books, Euclid treats of plane surfaces only.

PROP. XXVIII. THEOR.

GEN. ENUN.-If a straight line, falling upon two other straight lines, makes the exterior angle equal to the interior and opposite angle upon the same side of the line, or makes the interior angles upon the same side equal to two right angles, the two straight lines shall be parallel to one another.

PART. ENUN.—Let the st. line EF, falling upon the two st. lines AB, CD, make the exterior 2 EGB = the interior and opposite 2 GHD, upon the same side; or let it make the interior Zs on the same side, BGH, GHD, together = two rt. <$: then, in either case, AB is 1l to cd.

DEMONST.-1. Because the 2 AGH = vertical EGB (Prop. XV.) = L GHD (Hyp.), and they are alternate Zs; .. AB is ll to co. (Prop. XXVII.)

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B В

G

D

H

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2. Again, because the 28 AGH + BGH = two rt. 28 (Prop. XIII.) BGH + GHD (Hyp.); .. (by subtraction) the < AGH 2 GHD (Ax. 3), and these are alternate <$; .. AB is || to cd. (Prop. XXVII.)

Wherefore, if a st. line, &c.-Q. E. D.

PROP. XXIX. THEOR.

E E

A

B

GEN. ENUN.-If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another; and the exterior equal to the interior and opposite angle upon the same side ; and likewise the two interior angles upon the same side equal to two right angles.

PART. ENUN.-Let the st. line EF meet the ll st. lines AB, CD.

Then it makes the alternate ZS AGH, GHD =; also the exterior 2 EGB

interior and opposite 2 GHD, upon

the same side; and the two interior ZS BGH + GHD = two rt. Zs.

DEMONST.--1. For if the 2 Agu be not = Z GHD, one must be the >.

2. Suppose < AGH > than 2 GHD, and add to each Z BGH ; ..

ZS AGH + BGH are > the ZS. BGH + GHD (Ax. 4); but ZS AGH + BGH = two rt. ZS. (Prop. XIII.); .. zs BGH + GHD are < two rt. Zs.

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3. Now st. lines, which, when another line falls upon them, make the interior Zs upon the same side < two rt. 2$, will meet together, if continually produced (Ax. 12); .. the st. lines AB, CD will meet, if produced far enough.

4. But they never meet, because they are parallel (Def. 34); .. the 2 AGH is not unequal to the 2 GHD, i.e. the Z AGH =

- 2 GHD. 5. Again, the 2 AGH = Z EGB (Prop. XV.); .; Z EGB = Z GHD.

6. Add to each the < BGH; .. ZS EGB + BGH = < $ BGH + GHD (Ax. 2); but < $ EGB + BGH = two rt. Zs (Prop. XIII.); .. 28 BGH + GhD = two rt. 28 (Ax. 1).

Wherefore, if a st. line &c.-Q. E. D. This Proposition is converse to the two which precede. It depends upon the Twelfth Axiom, which, however, so far from being self-evident, is rather a Corollary to the Theorem which it is applied to demonstrate. This circumstance has given much trouble to mathematicians, and involves the doctrine of parallel lines in considerable obscurity. Without adverting to the different means which have been employed to get rid of the difficulty, suffice it to remark that the best way is to seek another solution of this Proposition, from which the proof of the Twelfth Axiom, or, more properly speaking, the Theorem which it involves, will readily be deduced. For this purpose we may proceed thus :

If the L AGH be not = GHD, let Agh be the > of the two. At the pt. g in the st. line EF make the 2 HGK = L GHD (Prop. XXIII.), and let k meet cd in K. Then the exterior / GAD, of the AKGH, is > than the interior opposite / HGK (Prop. XVI.); but (by Construction) the HGK LGAD, the less

the greater,

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A

-B

C

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K

F

which is impossible ; .. the Aga is not unequal to the LGHD, i. e. the AGH = L GHD; but the LAGH, &c. &c. &c. as before.

COR.-If a st. line EF meets two st. lines AB, CD, so as to make the two interior / $ BGH, ghd taken together less than two rt. 2$, these st. lines, being continually produced, will at length meet on that side of EF on which are the 28 less than two rt. 2$, i. e. towards B and D.

DEMONST.-1. For, if not, AB and cd are either || to one another, or they will meet on the other side of EF.

2. Now they are not || , for then the 48 BGH + GHD would be = to two rt. 18 (Prop. XXIX.); but they are not. (Hyp.)

3. Neither do they meet on the other side of EF; for, if so, the LS AGH, Ghc are two of a

E A, and .. they are < than two rt. ZS; but the four 48 AGH + BGH + GHD A+ GHC = four rt. (Prop. XIII.); and the LS BGH + GhD are < two rt. 48 (Hyp.); .. the 28 AGH + GHC are > two rt. 1 $.

4. .. AB and cp do not meet to. wards A and c: and they are not ll ; .. they will meet, if produced, towards B and D.- -Q. E. D.

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B

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D

F

PROP. XXX. THEOR.

GEN. ENUN.-Straight lines, which are parallel to the same straight line, are parallel to one another.

Part. Enun.—Let AB, co be both || to EF; then AB is || to co.

CONST. — Let the st. line GHK cut AB, EF, CD.

DEMONST.-1. Because ABCis || to EF, and GHK cuts

А.

B

E

K

them, ... the Z AGH = alternate 2 GHF. (Prop. XXIX.)

2. Again, because Er is || to cd, and GK cuts them, .. the exterior Lght = interior 2 GKD. (Prop. XXIX.)

3. .. the 2 AGK = Z GKD (Ax. 1), and they are alternate <8; .. AB is || to cd.

Wherefore, st. lines, &c.-Q. E. D. It is clear that the same reasoning may be carried on to any number of lines ; nor is it necessary that er should lie between AB and cd; and the student should prove the Theorem, when it is placed either above ab or below cd.

PROP. XXXI. THEOR.

E

A

F

any pt.

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B

D

с

Gen. Enun. — To draw a straight line through a given point parallel to a given straight line.

Part. Enun.—Let a be the gn. pt., and BC the gn. st. line. Then it is required to draw through A a st. line || to Bc.

Const.-In BC take D, and join AD (Post. l). At the pt. a, in the st. line AD, make the 2 DAE = ADC (Prop. XXIII.); and produce the st. line EA to F (Post. 2). Then EF is the line required.

DEMONST.-Because the st. line Ad, which meets the two st. lines EF, BC, makes the alternate ZS EAD, ADC = to one another, .. Er is ll to BC. (Prop. XXVII.)

Therefore the st. line EAF is drawn through

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