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IF two equal ftraight lines AC, BD on the fame fide of the straight line AB, make the interior angles CAB, ABD equal to one another; the ftraight line CD which joins their extremities, makes with them angles ACD, CDB equal to one another. Join AD, BC: and because in the triangles CAB, ABD, CA, AB, are equal to C DB, BA, and the angle CAB equal to ABD; the base CB is equal * to AD: and in the triangles ACD, BDC, AG, CD are equal to BD, DC, and the bafe AD is equal to BC; therefore the angle ACD is A equal to the angle BDC.

b

BOOK I.

a 4. 1.

B

b 8.1.

COR. Hence, if AC, BE make equal angles CAB, ABE with AB, but the angle BEC be greater than ACE; the ftraight line AC is greater than BE.

For, if AC were equal to BE, the angle BEC would be equal to ACE, but it is not. If AC be less than BE, make BF equal to AC, and join CF; therefore the angle BFC is equal to ACF; but BFC is greater than BEC; therefore BCF, and much c 16. 1. more BCE, is greater than BEC; and it is alfo lefs; which is impoffible: Wherefore BE must be lefs than AC.

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IF the quadrilateral figure BCDE, which has the angles at C, E right angles, have the fides BE, CD produced, and meeting in A; the fide BC is greater than the fide DE.

From C draw a CF perpendicular to AB: AF is greater than a 12. 3. AE. For if CF fall on E, the angle CEB would be equal to DEB, each being a right angle, the less to the greater; which is impoffible: and if AF be lefs than AE, and CE be joined, CFEwould be a triangle, of which the A angles CFE, CEF are together greater than two right angles; which is im

E T B

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poffible; therefore AF is greater than AE: and because CFB, CBF are lefs than two right angles, and CFB is a right angle, CBF is less than a right angle; therefore CB is greater than b 17. 1. CF. In the fame manner, it may be proved, that each of the c 19. 1. angles ACF, ADE is less than a right angle: and the angles

Nn 2

ADE,

d 13. 1.

BOOK I. ADE, EDC are equal to two right angles; therefore EDC is greater than a right angle: and because the angle DEF is equal to EFC, but EDC is greater than DCF; therefore CF is greate Cor. to 1.er than DE: but CB is greater than CF; much more, thereof them. fore, is CB, greater than DE.

a II. 1.

C 4. 1.

PROP. III.

IF from A, C two points of the ftraight line AC, perpen diculars AB, CD be drawn to BD; and the angle BAC be greater than a right angle; DCA is lefs than a right angle.

a

Draw AE at right angles to AC, and let it meet BD in E; and make EF equal to EĂ, and EG to EB, and join FG, and because AE, EB are equal to

FE, EG, and the angle AEB

b

to

b 15. 1. is equal to FEG; therefore
the angle EGF is equal
ABE, that is, to a right angle.
In the fame manner, if FG
meet AC in H, and HK be made
equal to HF, and HL to HA,
and KL be joined, it may be
proved, that KL is at right
angles to AC: and becaufe AEGH
is a quadrilateral, of which HAE,
HGE are right angles, and AE,
HG meet in F; therefore AH is
greater than EG or EB. In
the fame manner, it may be

d 2. of this.

K

H

M

D

fhown, that every perpendicular cuts off from AC and BD, a fegment not less than any of the former; therefore, if this be done continually, fome of the perpendiculars to AC shall meet CD: Let KL meet CD in M: and because CLM is a right € 17: angle, LCM or ACD is lefs than a right angle.

2 12. I.

e

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If the straight line AC meet the two ftraight lines AB, CD, and make the angle ACD a right angle, and CAB an acute angle; any number of perpendiculars to CD can be drawn from points in AB, all of them making acute angles with AB towards B.

From C draw a CE perpendicular to AB: it fhall fall towards B, for if it should fall on the other fide of A, the angle C13. 1. CAE would be obtufe, and therefore the angles CAE, AEC

would

a

b 17. I.

would be greater than two right angles, which is impoffible; Book I. therefore CE falls towards B: From E draw EF, perpendicular to CD; and it may be fhown, as before, that it falls towards D: and because CEB is a right angle, FEB is lefs than a right angle. In the same manner, may any number of perpendiculars to CD be drawn. on the fide of EF towards B; all of them making acute angles with AB towards B.

COR. Hence, if EF be perpendicular to CD, from the point E in AB; and if all the perpendiculars

B

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Ai

to CD that can be drawn from points in AE, make acute angles
with AB towards B, the angle FEB is an acute angle. For
any number of perpendiculars to CD, can be drawn from points
in AB nearer to B, than any that are on the other fide of E, all
making acute angles towards B;
fome of these points fhall therefore
be nearer to B, than the point E:
Let GH be perpendicular to CD,
from G nearer to B than the point
E, and making HGB an acute
angle; therefore HGE is greater than a
caufe EF, GH are perpendicular to CD,
than a right angle; therefore FEG is less

C

PROP. V.

E

G

B

F

H D

right angle: and be- c 13. 1.
and HGE is greater
than a right angle. d 3. of

If two straight lines AB, CD be at right angles to the fame : ftraight line EF; any ftraight line CA at right angles to one of thefe lines AB, is also at right angles to the other CD.

K C

ED

For the perpendiculars to AB that can be drawn from points in CE, cannot all make acute angles with CD towards D, because then FED would be an acute angle a ; but it is not let therefore GH be perpendicular to AB, and HGE not be an acute angle: But neither is it an obtuse

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this.

a Cor. 4.

of this.

FB

b 3. of this.

but it is not; therefore HG must be at right angles to CD. Make GK equal to GE, and HL to HF, and join KL, LG, GF :

and

C

C

`Book I. and because in the triangles GHL, GHF, LH, HG are equal to FH, HG, and the angles at H right angles; therefore the base € 4.1. GL is equal to GF, and the angle HGL equal to HGF, and GLH to GFH but the whole angle HGK is equal to HGE, therefore the remaining angles LGK, FGE are equal: and because in the triangles GKL, GEF, KG, GL are equal to EG, GF, and the angle KGL to EGF; the base LK is equal to EF, and the angles GKL, GLK to the angles GEF, GFE ; but GEF is a right angle, therefore GKL is a right angle; and because the angle GLH is equal to GFH, and GLK to GFE, the whole HLK is equal to the whole HFE, that is, to a right angle. In the fame manner, if fegments of AB, CD be taken equal to FL, EK, it may be proved, that the ftraight line joining their extremities is at right angles to AB, CD; and fo on : Therefore a ftraight line can be found on the other fide of AC, that is, at right angles to AB, CD: let this be KL: and if ACE, ACK d 13. 1. be not right angles, one of them is greater than a right angle: b3. of this. let this be ACK; therefore LKC is lefs than a right angle:

a 5.of this, b 10. 1. C 12. 1.

d 15. 1.

d

and it is also a right angle; which is impoffible: Therefore CA is at right angles to CD.

PRO P. VI.

IF a ftraight line EF meet two straight lines AB, CD which are at right angles to fome straight line AC; the alternate angles AEF, EFD fhall be equal to one another.

a

b

K B

If EF be at right angles to one of them, it is alfo at right angles to the other. But if not, bifect EF in G, and draw GH perpendicular to CD; it is also per. pendicular to AB : let it meet AB in K: and because the angles EGK, FGH are vertical, they are equal; and the angles at H, K are right angles; there are therefore two angles of the triangle EGK, equal to two of the triangle FGH; and the fides EG, GF oppofite to equal angles are equal; therefore the third angle KEG is equal to the third angle HFG.

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D

IF two ftraight lines AB, CD be cut by a third AC, so as to make the interior angles BAC, ACD on the fame fide of it, together less than two right angles; AB and CD being produced, hall meet one another towards the parts on which are the two angles, which are less than two right angles.

-Let

a

b

c

b 12. 1.

Let CAB be the leffer of the two angles BAC, ACD, if Book I. they be unequal; and bisect AC in E a, and draw ↳ EF perpendicular to AB, and from C draw CG perpendicular to EF, a 10. 1. and produce it to H: and because AB, GH are at right angles to FG, and AC meets them; the angle BAC is equal to ACG: c6.of this. to each of them add the angle ACH; then BAC, ACH are equal to ACG, ACH, that is, to two right angles: But BAC, ACD d 13. 1. are lefs than two right angles; therefore the angle ACH is greater than ACD; and CD falls between CH and AB. Make AK equal to AC, and join CK: and becaufe CK meets the two ftraight lines AB, CH, which are at right angles to FG; the angle AKC is equal to KCH: but the angle ACK is equal to

C

e

e 5. 1.

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AKC, because AK is equal to AC; therefore the angle ACK is equal to KCH. In like manner, if KL be made equal to KC, and CL joined, it may be proved, that it bifects the angle KCH; and fo on. And because the angles ACK, KCL are each of them greater than the angle DCH; and that the angle DCH taken fome number of times, is greater than ACD; much more shall the fame number of the angles ACK, KCL, &c. make an angle greater than ACD: let the angle thus made, which is greater than ACD, be ACM: and becaufe ACM is greater than ACD, CD falls within the angle ACM; and therefore, if it be produced, it must cut AB betwixt the points A and M.

Thus the 12th axiom is accurately demonftrated: But the tediousness of the procefs, and the alledged obfcurity of the ptopofition, may perhaps be thought fufficient reafons for affuming a more obvious property of ftraight lines: But when propofitions are not felf-evident, authors are not agreed about what makes one of them more obvious than another. We come to the knowledge of many properties of figures, by experience and obfervation, before we be able to deduce them from felf-evident principles and this is the reafon of our more readily affenting to, propofitions in which lines alone are concerned, than to those in which angles are concerned, for we are conftantly using the former, but make few obfervations on the latter. But one perfon may be better acquainted with one property, and another with another property; which makes it difficult to determine which of them is the most obvious: and this is the reafon why

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