Sidebilder
PDF
ePub
[merged small][ocr errors][merged small][merged small][ocr errors][merged small]

If two equal straight lines AC, BD on the same side of the traight line AB, make the interior angles CAB, ABD equal to one another; the straight line CD whichi 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:

a 4. I. and in the triangles ACD, BDC, AC, CD are equal to BD, DC, and the base AD is equal to BC; therefore the angle ACD is A

В equal to the angle BDC.

DS. 1. CoR. Hence, if AC, BE make cqual angles CAB, ABE with AB, but the angle BEC be greater than ACE; the straight 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 c than BEC, therefore BCF, and much €16.1. more BCE, is greater than BEC; and it is also leis ; which is impollible : Wherefore BE must be less than AG.

[blocks in formation]

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 side BC is greater than the fide DE.

From C draw a CF perpendicular to AB: AF is greater than a 12. g. 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 impossible: and if AF be less 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 impoffible; therefore AF is greater than AE: and because CFB, CBF are less 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. 3• angles ACF, ADE is less than a right angle; and the angles Nn 2

ADE,

Book I. ADE, EDC are equal d to two right angles; therefore EDC is m greater than a right angle: and because the angle CEF is equal d 13. 1.

to EFC, but EDC is greater than DCF; therefore CF is great, e Cor.to 1. er

e than DE: but CB is greater than CF; much more, thereof them. fore, is CB, greater than DE.

[blocks in formation]
[ocr errors]

to

If from A, C two points of the straight line AC, perpendiculars AB, CD be drawn to BD; and the angle BAC be

greater than a right angle; DCA is less than a right angle. 2 11. I. Draw a 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 15. 1. is equal to FEG; therefore C4. 1. the angle EGF is equal

ABE, that is, to a right angle.
In the same manner, if FG
meet AC in H, and HK be made
equal to HF, and HL to HA, H
and KL be joined, it may be
proved, that KL is at right

M
angles to AC: and because AEXH
is a quadrilateral, of which HAE,
HGE are right angles, and AE,

HG meet in F; therefore AH is d 2. of

greater d than EG or EB. In this.

the same manner,
fhown, that every perpendicular cuts off from AC and BD, a
segment not less than any of the former; therefore, if this be
done continually, some of the perpendiculars to AC Mall meet

CD: Let KL meet CD in M: and because CLM is a right € 17. 1. angle, LCM or ACD is less than a right angle.

it

may be

[blocks in formation]

If the straight line AC meet the two straight 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 shall fall to

wards B, for if it should fall on the other fide of A, the angle C 13. !. CAE would be obtuse and therefore the angles CAE, AEG

a 12. I.

would

[ocr errors]

would be greater than two right angles, which is impossible; Book I.
therefore CE falls towards B: From E draw · EF, perpendi-
cular to CD; and it

may
be shown,

b 17. I.
as before, that it falls towards D:
and because CEB is a right angle,
FEB is less than a right angle. In
the same manner, may any number
of perpendiculars to CD be drawn

B
on the side of EF towards B; all
of them making acute angles with
AB towards B.

Cor. Hence, if EF be perpendi-
cular to CD, from the point E in
AB; and if all the perpendiculars
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
ang 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;
some of these points thall therefore

B
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 right angle : and be- c 13. 1.
cause EF, GH are perpendicular to CD, and HGE is greater
than a right angle; therefore FEG is less than a right angle. d 3. of

this.

[blocks in formation]

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

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

a Cor. 4. not: let therefore GH be

of this. perpendicular to AB, and HGE not be an acute angle: But neither is it an obtuse angle, because then GEF

11

FB would be an acute angle b;

b but it is not; therefore HG must be at right angles to

3. of CD.

this. Make GK equal to GE, and HL to HF, and join KL, LG, GF:

and

[ocr errors]

Book I. and because in the triangles GHL, GHF, LH, HG are equal to

NFH, HG, and the angles at H right angles; therefore the base C 4. 1. GL is equal to GF, and he 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 10 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 20-le. ln the same manner, if segments of AB, CD be taken equal to FL, EK, it may be proved, that the straight line joining their extremities is at right angles to AB, CD; and so on : Therefore a straight line can be found on the other side 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 less than a right angle:

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

с

PRO P. VI.
If a straight line EF meet two straight lines AB, CD which
are at right angles to some straight line AC; the alternate angles
AEF, EFD shall be equal to one another.

If EF be at right angles to one of them, it is also at right a s.of this, angles a to the other. But if not, bisect b EF in G, and draw b 10. 1. GH perpendicular to CD, it is also per. pendicular to AB 2 : let it meet AB in

K 123 K: and because the angles EGK, FGH el d 15. 1. are vertical, they are equal d; 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 sides EG,
GF opposite to equal angles are equal; therefore the third angle
KEG is equal to the third angle HFG.

CI2. 1.

[blocks in formation]

1F two straight lines AB, CD be cut by a third AC, so as to. make the interior angles BAC, ACD on the same fide of it, to. gether less than two right angles ; AB and CD being produced, Thall meet one another towards the parts on which are the two angles, which are less than two right angles.

Let

Let CAB be the lesser of the two angles BAC, ACD, if Book I. they be unequal; and bisect AC in E a, and draw 6 EF perpendicular to AB, and from C draw 5 CG perpendicular to EF, a 10. 1. and produce it to H: and because AB, GH are at right angles b 12. 1. 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 less 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 because CK meets the two straight lines AB, CH, which are at right angles to FG; the angle AKC is equal to KCH: but the angle ACK is equal to e 5. 1.

[ocr errors]

B

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 bisects the angle KCH;
and so on. And because the angles ACK, KCL are each of thern
greater than the angle DCH; and that the angle DCH taken
fome number of times, is greater than ACD; much more shall
the same 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 because ACM is greater than ACD;
CD falls within the angle ACM; and therefore, if it be pro-
duced, it mult cut AB betwixt the points A and M.

Thus the 12th axiom is accurately demonstrated : But the tediousness of the process, and the alledged obfcurity of the proposition, may perhaps be thought sufficient reasons for assuming a more obvious property of straight lines : But when propofitions are not self-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 atid obfervation, before we be able to deduce them from self-evident principles : and this is the reason of our more readily afsenting to propofitions in which lines alone are concerned, than to thole in which angles are concerned, for we are constantly using the former, but make few obfervations on the latter. But one person may be better acquainied 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 reason why

the

« ForrigeFortsett »