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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 LEGB = the interior and opposite

GHD, upon the same side; or let it make the interiors on the same side,

=

BGH, GHD, together two rt. 4s: then, in either case, AB is to CD.

DEMONST.-1. Because

the AGH = vertical 4

A

C

E

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EGB (Prop. XV.) = GHD (Hyp.), and they are alternates; .. AB is | to CD. (Prop. XXVII.)

2. Again, because the

S AGH + BGH =

= BGH GHD

two rt. 4s (Prop. XIII.)
(Hyp.); .. (by subtraction) the

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GHD (Ax. 3), and these are alternate <s;

.. AB is to CD. (Prop. XXVII.) Wherefore, if a st. line, &c.-Q. E. D.

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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 st. lines AB, CD. Then it makes the alter

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DEMONST.-1. For if the AGH be not =
GHD, one must be the >.

S

2. Suppose AGH> than 2 GHD, and add to each BGH; ZS AGH + BGH are > the s BGH + GHD (Ax. 4); but ≤s AGH + BGH = two rt. ≤s (Prop. XIII.); .'. ▲s BGH GHD are < two rt. Zs.

3. Now st. lines, which, when another line falls upon them, make the interior s upon the same side < two rt. <s, 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 ▲ AGH is not unequal to the GHD, i. e. the Z AGH =

5. Again, the AGH = XV.); .. ▲ EGB = ▲ GHD.

S

GHD.

EGB (Prop.

6. Add to each the BGH; .. s EGB + BGH = ≤s BGH + GHD (Ax. 2); but ≤3 EGB + BGH = two rt. ≤s (Prop. XIII.); .. 2o BGH + GHD = two rt. ≤s (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:

A

E

B

If the AGH be not = GHD, let AGH be the > of the two. At the pt. G in the st. line EF make the HGK = Z GHD (Prop. XXIII.), and let GK meet CD in K. Then the exterior GHD, of the ▲ KGH, is than the interior opposite HGK (Prop. XVI.); but K (by Construction) the GHD, the less

=

=

HGK
the greater,

C

-D

H

which is impossible; .. the ▲ AGH is not unequal to the GHD, i. e. the GHD; but the AGH, &c. &c. &c. as before.

AGH

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. 4s, these st. lines, being continually produced, will at length meet on that side of EF on which are the less than two rt. Zs, 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 S BGH + GHD would be = to two rt. 4 (Prop. XXIX.); but they are

not. (Hyp.)

3. Neither do they meet on the other side of EF; for, if so, the SAGH, GHC are two 4s of a

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A, and.. they are < than two rt. 2s;
but the four 8 AGH + BGH + GHD A
+ GHC = four rt. 4 (Prop. XIII.);
and the BGH + GHD are < two rt.
< (Hyp.); .. the ▲ AGH + GHC
are two rt. s.

4. .. AB and CD do not meet towards A and c: and they are not ;

E

H

C

F

G

... they will meet, if produced, towards в and D.-Q. E. D.

PROP. XXX. THEOR.

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

PART. ENUN.-Let AB, CD be both to EF; then AB is || to CD.

CONST.-Let the st. line

GHK cut AB, ef, cd.

DEMONST.-1. Because AB is to EF, and GHK cuts

A

B

H

E

K

C

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

2. Again, because EF is to CD, and GK cuts them, .. the exterior ▲ GHF = interior GKD. (Prop. XXIX.)

3... the

AGK = 2 GKD (Ax. 1), and they are alternate s; .. 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 EF 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.

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 вc the gn. st. line. Then it is required to draw through A a st. line to BC.

E

B

D

A

F

с

CONST.-In BC take any pt. D, and join AD (Post. 1). At the pt. A, in the st. line AD, make the 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 s EAD, ADC = to one another, .. Er is to BC. (Prop. XXVII.)

Therefore the st. line EAF is drawn through

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