Book I.

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.

Let the line CD turn about the point C till it coincide with CE, a part of the fide CA, and have defcribed the exterior angle DCE or DCA. Let it then be carried along the line CA till it be in the pofition AF, 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 pofition BG, and, by turning about B, let it defcribe the angle

F

B

GBC, fo as to coincide with a part of BC.
Laftly, Let it be
carried along BC, fo as to coincide with CD, its firft pofition.
Then, because the line CD has turned about one of its extre-
mities till it has come into the pofition CD again, it has by
the Cor. to the above Def. described angles which are toge-
ther equal to four right angles; but the angles which it has
defcribed 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 fides, the propofition is demonftrated juft as in the case of a triangle. Therefore all the exterior angies of any rectilineal figure are together equal to four right angles. QE. D.

Book I.

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 fix 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 oppofite, or the angle DCA is equal to the engles CAB, ABC. For the angles CAB, ABC, BCA are equal to two right angles; and the angles ACD, ACB are alfo (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 fides, wanting four. For all the angles exterior and interior are equal to twice as many right angles as the figure has fides; but the exterior are equal to four right angles; therefore the interior are equal to twice as many right angles as the figure has fides, wanting four.

PROP. II.

Two ftraight lines, which make with a third line the interior angles on the fame fide of it lefs than two right angles, will meet on that fide, if produced far enough.

Let the ftraight lines AB, CD, make with AC the two angles BAC, DCA lefs 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 alfo equal (5. 1.); but the exterior angle HAC is equal to the two interior and oppofite 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

and is bifected by the line CF. In the fame manner, if FG Book I. be taken equal to FC, and if CG be drawn, it may be shewn that CG bifects the angle ACE, and fo 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 lefs than the given angle DCE. Let GCE be the

angle, whofe half ECK is lefs than DCE, then a ftraight line CK is found, which falls between CD and CE, but neverthelefs 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.

PROP.

Prop. 1. 1. Sup. The reference to this propofition involves nothing inconfiftent with good reafoning, as the demonftration of it does not depend on any thing that has gone before, fo that it may be introduced in any part of the Ele

ments.

Book I.

PROP. III.

29. 1. Euclid.

If a straight line fall on two parallel ftraight lines, it makes the alternate angles equal to one another; the exterior equal to the interior and oppofite on the fame fide; and likewife the two interior angles, on the fame fide equal to two right angles.

Let the ftraight line EF fall on the parallel ftraight lines AB, CD; the alternate angles AGH, GHD are equal, the exterior angle EGB is equal to the interior and oppofite 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 AGH, HGB are greater than the angles DHG, HGB But AGH, HGB are equal to two right angles, (13.); therefore BGH, GHD are lefs than two right angles, and therefore the lines AB, CD will meet, by the last propofition, if produced toward B and D. But they do not meet, for they are parallel by hy pothefis, 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 alfo fhewn to be equal to GHD, therefore EGB and GHD are equal. Laftly, to each of the equal angles EGB, GHD add the angle BGH, then the two EG, BGH are equal to the two DHG, BGH. But EGB, BGH are equal to two right angles, (13. 1.), therefore BGH, GHD are alfo equal to two right angles. Therefore, &c. Q. E. D.

The

Book I.

The following propofition is placed here, because it is more connected with the first Book than with any other. It is ufeful for explaining the nature of Hadley's fextant; and though involved in the explanations usually given of that inftrument, it has not, I believe, been hitherto confidered as a diftinct Geometric Propofition, though very well entitled to be fo on account of its fimplicity and elegance, as well as its utility.

THEOREM.

If an exterior angle of a triangle be bifected, and also one of the interior and oppofite, the angle contained by the bifecting lines is equal to half the other interior and oppofite angle of the triangle.

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

The lines CE, BE will meet; for fince 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 lefs than two right angles, and therefore the lines CE, BE muft meet on the fame fide 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