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*INTERC. 9.

In BC take any point as D, and join* AD; and through the Cor. +Interc.14. pointsB,C,A, let a plane bet made to pass, which (because BC is

a straight line) will also have the point D in it, and consequently

also the straight line DA. In this plane, on the straight line AD #1.22.Cor.2. and at the point A in it, makes the angle DAE equal to the angle *Interc.12. ADC; and prolong* EA to F. Cor. 6.

Because the straight line AD meets the two straight lines

BC and EF which are both in the same plane, and the alternate
+ Constr. angles EAD and ADC aret equal to one another, EF and BC are
#1.27.Cor.l,

parallel. Therefore through the point A is drawn a straight line
EAF, parallel to BC. Which was to be done.

And by parity of reasoning, the like may be done in every
other instance.

12.

PROPOSITION XXVIII A.

THEOREM.-If from the ends of a given straight line (which

shall for distinction be called the base], two straight lines equal
to one another be drawn towards the same side, making equal
interior angles at the base, each less than the sum of two right
angles, but the two equal straight lines do not meet ; and the
extremities of the two equal straight lines be joined ; there shall
be formed a quadrilateral figure, on that side of the base on
which are the angles each less than the sum of two right
angles; of which the angles opposite to the base shall be
equal to one another, and each less than the sum of two right
angles.

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Let AB be the given straight line, and from its ends A and
B let there be drawn the two equal straight lines AD and
BC towards the

Z
same side, mak-
2

2
D F

D
ing equal inte-

DF
rior angles BAD
and ABC with

A E

А

в А E
AB the base,
each less than the sum of two right angles, but AD and
BC do not meet (as might happen where the angles BAD

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Cor.

*Interc. 9. and ABC are less than right angles); and let DC be* joined.

There shall be formed a quadrilateral figure ABCD, on the side of
AB on which are the angles BAD and ABC ; of which the angles
ADC, BCD, which are opposite to AB, shall be equal to one

another, and each less than the sum of two right angles. + 1.9. Bisectt AB, in E; and from E drawf a straight line EZ of 1 1. 10.

unlimited length, at right angles to AB, on the side of it on which are the angles BAD and ABC. Join also ED; EC.

Because at the point B the straight lines BE and BC make an Нур. .

angle EBC that is* less than the sum of two right angles, and

E and C which are points in these two straight lines are joined ; +113.Cor.3. there will be formedt a triangle EBC, on the side on which is

the angle EBC that is less than the sum of two right angles.

In the same way EAD will be a triangle. And because the 11.12.Cor.1. angles BEC, CED, AED aref together equal to two right

angles, CED is less than the sum of two right angles. Therefore *1.13.Cor.3. the straight line DC, which joins the points D and C, will*

form a triangle with the straight lines ED and EC on the side on which is the angle CED. Wherefore DC will lie wholly on the side of AB on which are the angles BAD and ABC that are each less than the sum of two right angles; and ABCD will be

a quadrilateral figure on the same side. And because in the tri. + Constr.

angles EDA and ECB, AEt and AD1 are equal to BE and BC | Hyp. Нур.

respectively, and the angle EAD is* equal to the angle EBC, I. 4.

ED ist equal to EC, and the angle AED to BEC, and the angle EDA to ECB. And because BEC, CED, AED are together equal to two right angles, BEC and AED are together less than two right angles; and because they are also equal to one another, they are each less than one right angle, and less than AEZ or BEZ; where

fore EC, ED will fall on different sides of EZ. And because the 11. 11. angle AEZ ist equal to BEZ, and AED to BEC, the remaining "Interc. 1. angle DEZ* is equal to the remaining angle CEZ. Wherefore the Cor. 7.

angle CED is bisected by the straight line EZ; and because in the angles DEF, CEF, the sides ED and EF are equal to the sides

EC and EF respectively, and the angle DEF has been shown to #. 4.

be equal to the angle CEF, FD ist equal to FC; and the angle

EFD is equal to EFC, and because they are adjacent angles 11.Nom.37. they are right angles ; and the angle EDF is equal to ECF. But

the angle EDA has been shown to be equal to ECB; therefore *INTERC. 1. the sum of the angles EDF and EDA is* equal to the sum Cor. 5.

of the angles ECF and ECB; or the angle ADC is equal to the angle BCD. Also these are each less than the sum of two

right angles. For if they were each equal to two right angles, *1.13.Cor.4. ADCB would be* one straight line, and two straight lines ADCB + INTERC.10. and AB would inclose a space, which ist impossible ; and if they Cor. l.

were yet greater [that is, if DA and CB made with DC equal angles each less than two right angles on the side towards Z] the straight line joining the extremities A and B must (as before shown) lie on the side of DC which is towards Z ; and it does not, for it is on the other side. And since they are each neither equal to the sum of two right angles nor greater, they are less.

And by parity of reasoning, the like may be proved of all other straight lines under the same conditions. Wherefore, universally, if from the ends of a given straight line &c. Which was to be demonstrated.

COR. 1. The side of the quadrilateral figure which is opposite to the base, is bisected at right angles by the perpendicular drawn from the middle of the base ; and is parallel to the base.

For it has been shown that EF bisects DC at right angles. Constr. Wherefore, because EFC is a right angle, and AEF also is! a * I. 11. right angle, they are* equal to one another. And they are +1.27.Cor.), alternate angles; therefore DC and AB aret parallel.

Cor. 2. If through any point in EF as G, a straight line of unlimited length both ways (as WX) be drawn at right angles to EF; it shall cut the straight lines AD and BC between their extremities, and make with them angles GKA, GIB on the side towards AB the base, each greater than the angle ACB which is between the straight lines drawn from C to the two extremities of the base.

For WX cannot meet either AB or CD; because, since the

alternate angles DFG, FGX, and WGE, GEB, are right angles, 1. 11. and consequently equal to one another, it will be* parallel to them *1.27.Cor.1. both. Therefore it will cut AD and BC between their extremi

ties; for if it did not, it must meet either AB or CD. Let it

cut them in I and K; and because GIB is the exterior angle of + 1. 16.

the triangle HIC, it ist greater than the interior and opposite HCI or ACB.

Z

D

F

W

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X

K

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A.

E

B

CoR. 3. If a straight line WX of unlimited length both ways, be made to pass through the point E at right angles to EZ, and afterwards be moved along EZ keeping ever at right angles to it, till it pass through the point F; it will never cease to cụt the straight lines AD, BC, but will always cut them and make with each of them, on the side towards AB, an angle greater than ACB.

For (by Cor. 2 above) it will make such angles with them, in all and every of the situations through which it moves.

NOMENCLATURE.—A quadrilateral figure as above, viz. of which two opposite sides are equal, and the two interior angles made by them with a side between them are equal; is called a tessera.

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THEOREM.-If two equal straight lines make an angle less than

the sum of two right angles, and at the outward extremity of each be added another straight line equal to the first, having its extremity terminated in the extremity of the other, and making with it an angle equal to the first-mentioned angle and on the same side ; and at the outward extremity of each of these last, be added another equal straight line as before, and so on continually; and the extremities of every two equal straight lines so successively added at one and the same time at the two ends of the series, be joined by a straight line or chord; each of these chords shall make the angles at the two cusps, where it meets the series, equal to one another; and the several chords shall in succession make greater and greater angles at the cusp, each than the preceding.

Let AB, AC be two equal straight lines, making an angle BAC less than the sum of two right angles ; and at B, C, let there be added the straight lines BD, CE, each equal to AB or AC, and making the angles ABD, ACE, each equal to BAC; and let BC, DE be* joined. And in like manner let there be added DF and EG, each equal to AB or AC, and

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M

X

B

gles, the

making the angles BDF,

z
CEG, each equal to BAC,
and let FG be joined ; and
so on successively. The
angles ABC, ACB, which
are at the two ends of the
straight line BC, shall be
equal to one another; and
the angles BDE, CED,
which are at the two cusps
of the figure DBACE,
shall also be equal to one

1
another; and the angles

w
DFG, EGF, which are
at the two cusps of the
figure FDBACEG, shall
be equal to one another ; &c. Also the angle BDE shall be

greater than the angle ABC, and DFG than BDE; and so on.
*INTERC. 9. Join* BE, DG, FI, &c.
Cor.
+Hyp.

Because BAC ist less than the sum of right 11.13.Cor.3. straight line which joins B and C willt form a triangle. And

because in the triangle BAC the sides AC and AB are equal, the # 1.5. + Hyp.

angle ABC is* equal to ACB. From the equalt angles ABD,

ACE (which are each less than the sum of two right angles), take INTERC. 1. away the equal angles ABC, ACB, and there remain the equal Cor.7.

angles CBD, BCE, each less than the sum of two right angles. Нур. : Also, the sides BD, CE are* equal to one another; wherefore +1. 28 A.

DBCE ist a tessera, and the angles BDE, CED equalf to one Nom. 1. 28 A. another. And in the same way may be shown that FDEG, HFGI,

&c. are likewise tesseras. Again, because at the ends of the

base AC are drawn* two equal straight lines AB, CE, making with + Hyp:

AC the equalt angles CAB, ACE, each of which is less than the
sum of two right angles; BACE is a tessera, and the angles
ABE, CEB are equal to one another. And because the angles

ABE, CEB are equal to one another, and the angles ABD, CEG | Hyp. aref equal to one another, the remaining angle EBD will be*

equal to the remaining angle BEG; wherefore, because BD, EG
Cor. 7.
+ Hyp: aret equal to one another, DBEG is also a tessera ; and in the

same manner may be shown that FDGI, &c. are likewise
tesseras. And because the angle ABE is greater than ABC, and

.

• Hyp:

1

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