Sidebilder
PDF
ePub

Though this method of treating the subject is strictly demonstrative, yet, as the reasoning in the first of the two preceding demonstrations is not perhaps sufficiently simple to be apprehended by those just entering on mathematical studies, I shall submit to the reader another method, not liable to the same objection, which I know, from experience, to be of use in explaining the Elements. It proceeds, like that of the French geometer, by demonstrating, in the first place, that the angles of any triangle are together equal to two right angles and deducing from thence, that two lines, which make with a third line the interior angles, less than two right angles, must meet if produced. The reasoning used to demonstrate the first of these propositions may be objected to by some as involving the idea of motion, and the transference of a line from one place to another. This, however, is no more than Euclid has done himself on some occasions: and when it furnishes so short a road to the truth as in the present instance, and does not impair the evidence of the conclusion, it seems to be in no respect inconsistent with the utmost rigour of demonstration. It is of importance in explaining the Elements of Science, to connect truths by the shortest chain possible; and till that is done, we can never consider them as being placed in their natural order. The reasoning in the first of the following propositions is so simple, that it seems hardly susceptible of abbreviation, and it has the advantage of connecting immediately two truths so much alike, that one might conclude, even from the bare enunciations, that they are but different cases of the same general theorem, viz. That all the angles about a point, and all the exterior angles of any rectilineal figure, are constantly of the same magnitude, and equal to four right angles.

DEFINITION.

Ir, while one extremity of a straight line remains fixed at A, the line itself turns about that point from the position AB to the position AC, it is said to describe the angle BAC contained by the lines AB and AC.

[blocks in formation]

COR. If a line turn about a point from the position AB till it come into the position AB again, it describes angles which are together equal to four right angles. This is evident from the second Cor. to the 15th.

PROP. I.

All the exterior angles of any rectilineal figure are together equal to four right angles.

1. Let the rectilineal figure be the triangle ABC, of which the exterior angles are DCA, FAB, GBC; these angles are together equal to four right angles.

F

Let the line CD, placed in the direction of BC produced, turn about the point C till it coincide with CE, a part of the side CA, and have described the exterior angle DCE or DCA. Let it then be carried along the line CA, till it be in the position AF, that is in the direction of CA produced, and the point A remaining fixed, let it turn about A till it describe the angle FAB, and coincide with a part of the line AB. Let it next be carried along AB till it come into the position BG, and by turning about B, let it describe the angle GBC, so as to coincide with a part of BC. Lastly, Let it be carried along BC till it coincide with CD, its first position. Then, because the line CD has turned about one of its extremities till it has come.

E

B

G

D

into the position CD again, it has by the corollary to the above definition described angles which are together equal to four right angles; but the angles which it has described are the three exterior angles of the triangle ABC, therefore the exterior angles of the triangle ABC are equal to four right angles.

2. If the rectilineal figure have any number of sides, the proposition is demonstrated just as in the case of a triangle. Therefore all the exterior angles of any rectilineal figure are together equal to four right angles. Q. E. D.

COR. 1. Hence, all the interior angles of any triangle are equal to two right angles. For all the angles of the triangle, both exterior and interior, are equal to six right angles, and the exterior being equal to four right angles, the interior are equal to two right angles.

COR. 2. An exterior angle of any triangle is equal to the two interior and opposite, or the angle DCA is equal to the angles CAB, ABC. For the angles CAB, ABC, BCA are equal to two right angles; and the angles ACD, ACB are also (13. 1.) equal to two right angles; therefore the three angles CAB, ABC, BCA are equal to the two ACD, ACB and taking ACB from both, the angle ACD is equal to the two angles CAB, ABC.

COR. 3. The interior angles of any rectilineal figure are equal to twice as many right angles as the figure has sides, wanting four. For all the angles exterior and interior are equal to twice as many right angles as the figure has sides; but the exterior are equal to four right angles; therefore the interior are equal to twice as many right angles as the figure has sides, wanting four.

PROP. II.

Two straight lines, which make with a third line the interior angles on the same side of it less than two right angles, will meet on that side, if produced far enough.

Let the straight lines AB, CD, make with AC the two angles BAC, DCA less than two right angles; AB and CD will meet if produced toward B and D.

In AB take AF=AC; join CF, produce BA to H, and through C draw CE, making the angle ACE equal to the angle CAH.

Because AC is equal to AF, the angles AFC, ACF are also equal (5. 1.); but the exterior angle HAC is equal to the two interior and opposite angles ACF, AFC, and therefore it is double of either of them, as of ACF. Now ACE is equal to HAC by construction, therefore ACE is double of ACF, and is bisected by the line CF. In the same manner, if FG be taken equal to FC, and if CG be drawn, it may be shewn that CG bisects the angle ACE, and so on continually. But if from a magnitude, as the angle ACE, there be taken its half, and from ` the remainder FCE its half FCG, and from the remainder GCE its half, &c. a remainder will at length be found less than the given angle DCE.*

[ocr errors]
[merged small][merged small][ocr errors][ocr errors][merged small]

Let GCE be the angle, whose half ECK is less than DCE, then a straight line CK is found, which falls between CD and CE, but nevertheless meets the line AB in K. Therefore CD, if produced, must meet AB in a point between G and K. Therefore, &c. Q. E. D.

This demonstration is indirect; but this proposition, if the definition of parallels were changed, as suggested at p. 302, would not be necessary; and the proof, that lines equally inclined to any one line must be so to every line, would follow directly from the angles of a triangle being equal to two right angles. The doctrine of parallel lines would in this manner be freed from all difficulty.

PROP. III. 29. 1. Euclid.

If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another; the exterior equal to the interior

* Prop. 1 1. Sup. The reference to this proposition involves nothing inconsistent with good reasoning, as the demonstration of it does not depend on any thing that has gone before, so that it may be introduced in any part of the Elements.

and opposite on the same side; and likewise the two interior angles, on the same side equal to two right angles.

Let the straight line EF fall

on the parallel straight lines
AB, CD; the alternate angles
AGH, GHD are equal, the
exterior angle EGB is equal A-
to the interior and opposite
GHD; and the two interior
angles BGH, GHD are equal

to two right angles.

For if AGH be not equal to GHD, let it be greater, then adding BGH to both, the angles

[blocks in formation]

C

[blocks in formation]

AGH, HGB are greater than the angles DHG, HGB. But AGH, HGB are equal to two right angles, (13.); therefore BGH, GHD are less than two right angles, and therefore the lines AB, CD will meet, by the last proposition, if produced toward B and D. But they do not meet, for they are parallel by hypothesis, and therefore the angles AGH, GHD are not unequal, that is, they are equal to one another.

Now the angle AGH is equal to EGB, because these are vertical, and it has been also shewn to be equal to GHD, therefore EGB and GHD are equal. Lastly, to each of the equal angles EGB, GHD add the angle BGH, then the two EGB, BGH are equal to the two DHG, BGH. But EGB, BGH are equal to two right angles, (13. 1.), therefore BGH, GHD are also equal to two right angles. Therefore, &c. Q. E. D.

The following proposition is placed here, because it is more connected with the First Book than with any other. It is useful for explaining the nature of Hadley's sextant; and though involved in the explanations usually given of that instrument, it has not, I believe, been hitherto considered as a distinct Geometric Proposition, though very well entitled to be so on account of its simplicity and elegance, as well as its utility.

THEOREM.

If an exterior angle of a triangle be bisected, and also one of the interior and opposite, the angle contained by the bisecting lines is equal to half the other interior and opposite angle of the triangle.

Let the exterior angle ACD of the triangle ABC be bisected by the straight line CE, and the interior and opposite ABC by the straight line BE, the angle BEC is equal to half the angle BAC.

The lines CE, BE will meet; for since the angle ACD is greater than ABC, the half of ACD is greater than the half of ABC, that is, ECD

is greater than EBC; add ECB to both, and the two angles ECD, ECB are greater than EBC, ECB. But ECD, ECB are equal to two right angles; therefore ECB,EBC, are less than two right angles, and therefore the lines CE, BE must meet on the

B

E

A

D

same side of BC on which the triangle ABC is. Let them meet in E. Because DCE is the exterior angle of the triangle BCE, it is equal to the two angles CBE, BEC, and therefore twice the angle DCE, that is, the angle DCA is equal to twice the angles CBE, and BEC. But twice the angles CBE is equal to the angle ABC, therefore the angle DAC is equal to the angle ABC, together with twice the angle BEC; and the same angle DCA being the exterior angle of the triangle ABC, is equal to the two angles ABC, CAB, wherefore the two angles ABC, CAB are equal to ABC and twice BEC. Therefore, taking away ABC from both, there remains the angle CAB equal to twice the angle BEC, or BEC equal to the half of BAC. Therefore, &c. Q. E. D.

BOOK II.

THE Demonstrations of this Book are no otherwise changed than by introducing into them some characters similar to those of Algebra, which is always of great use where the reasoning turns on the addition or subtraction of rectangles. To Euclid's demonstrations, others are sometimes added, serving to deduce the propositions from the fourth, without the assistance of a diagram.

PROP. A and B.

These Theorems are added on account of their great use in geometry, and their close connection with the other propositions which are the subject of this Book. Prop. A is an extension of the 9th and 10th.

BOOK III.

DEFINITIONS.

The definition which Euclid makes the first of this Book is that of equal circles, which he defines to be "those of which the diameters

« ForrigeFortsett »