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BA cannot fall on the side of ED which is towards H. And in the same way may be shown, that it cannot fall on the side of ED which is towards F; and because it does not fall on either side of ED, it falls upon it, and the angle ABC coincides with the angle DEF, and is equal to it.

And by parity of reasoning, the like may be proved in every other instance, Wherefore, universally, right angles are equal to one another. Which was to be demonstrated.

PROPOSITION XII.
THEOREM.The angles which one straight line makes with

another on the one side of it, are together equal to two right
angles.

Let the straight line AB make with CD, on the one side of it, the angles ABC, ABD. These are together

А equal to two right angles.

First Case. If the angles ABC, ABD are I.Nom.37. equal to one another, each of them is* a right angle, and consequently they are together D

B C equal to two right angles.

Second Case. If the angles ABC, ABD are not equal to one f 1. 10. another, from the point B drawt BE at right angles to CD; therefore the angles EBD, EBC are two right

E

A angles.

Because the angle ABD is equal to the angles
ABE and EBD together, if to each of these

D
INTERC. 1. equals be added the angle ABC, the sums willi

B Cor. 4.

be equal; that is to say, the sum of the angles ABC and ABD, will be equal to the sum of the angles ABC, ABE, EBD. Again, because the angle EBC is equal to the angles ABC and ABE together, if to each of these equals be added the angle EBD, the sums will be equal; that is to say, the sum of the angles EBC, and EBD, will be equal to the sum of the angles ABC, ABE, EBD. But the sum of ABC and ABD was shown to be also equal to

the sum of ABC, ABE, EBD; therefore the sum of ABC and *INTERC. 1. ABD is* equal to the sum of EBC and EBD. But EBC, EBD + Constr. aref two right angles ; therefore the sum of the angles ABC and I INTERC. 1. ABD ist also equal to two right angles. Cor. 1.

And by parity of reasoning, the like may be proved in every

other instance. Wherefore, universally, the angles which one straight line makes with another on the one side of it, &c. Which was to be demonstrated.

Cor. 1. All the angles, however many, on the one side of a straight line, at the same point and in the same plane, are together equal to two right angles.

For their sum is equal to the sum of the two angles made by any one of the straight lines that fall upon the other, on the one side.

Cor. 2. All the angles made by any number of straight lines proceeding from one point and in the same plane, are together equal to four right angles.

For if any one of the straight lines be prolonged on the other side of the angular point, the angles on each side of this line will (by Cor. 1 above) be together equal to two right angles, and those on both sides will be equal to four.

PROPOSITION XIII.
THEOREM.-If, at a point in a straight line, two other straight

lines, on the opposite sides of it, make the adjacent angles
together equal to two right angles; those two other straight lines
shall be in one and the same straight line.

At the point B in the straight line AB, let the straight lines BD, BC, on the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles. BC is in the same straight line with DB.

D B C 'INTERC.12.

Prolong* DB on the side of B. If then it be disputed that the Cor.6. straight line so added shall coincide with BC, let it be assumed

that it falls on one side of it, as BE.

Therefore, because the straight line AB makes angles with the

straight line DBE on the one side of it, the angles ABD, ABE + 1. 12.

must be togethert equal to two right angles. But the angles ABD, | Hyp.

ABC are together equal to two right angles; wherefore the INTERC. 1. angles ABD, ABE are together equal* to ABD, ABC. Take

away the common angle ABD, and the remaining angle ABE +Interc. 1. must bet equal to the remaining angle ABC, the less to the greater; Cor. 7.

which is impossible. DB prolonged, therefore, cannot fall on the side of BC which is towards A; and in the same way may be shown, that it cannot fall on the other side ; and because it does

F

not fall on either side of it, it falls upon it, and consequently BC is in one and the same straight line with DB.

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

CoR. 1. Any straight line making with another an angle less or greater than the sum of two right angles, is not in the same straight line with it.

For let BD, BE be two straight lines making an angle DBE

less than the sum of two right angles. From B draw any straight "INTERC.12. line BA between BD and BE; and prolong* DB to C. Because Cor. 6.

ABD, ABC (by Prop. XIII above) will be together equal to two

right angles, and ABD, ABE (which are equal to DBE) are + Hyp. togethert less than two right angles, ABD and ABC will be IINTERC.I. Cor. 2. together greater than ABD and ABE, and by taking away the *INTERC. 1. common angle ABD, the angle ABC will be* greater than ABE Cor. 8. + Constr. consequently BE does not coincide with BC. And because BC ist

in the same straight line with DB, and BE does not coincide with it, BE is not in the same straight line with DB. Also if BD,

BE make an angle greater than two right angles ; because all the 31.12.Cor.2. angles at the point B are together equal to four right angles, and

the angle on one side of BD and BE is greater than two right angles, the angle on the other side is less ; wherefore, as before, BD and BE are not in the same straight line.

;

CoR. 2. Any two straight lines which meet and make an angle DBE less than the sum of two right angles, being prolonged shall cut one another.

For because the angle ABC (as in Cor. 1 above) is greater than ABE, BC (which is the prolongation of DB) will fall on the side of BE which is most remote from A. And in like manner may be shown that EB being prolonged will fall on the side of BD which is most remote from A. That is, DB and EB being prolonged will cut one another.

Cor. 3. If from any point (as B) in a straight line DC, be drawn another straight line BE making with BD an angle that is less than the sum of two right angles; and two other points in these straight lines (as D, E,) be joined ; there shall be formed a

triangle, on that side of DC on which is the angle that is less than the sum of two right angles ; and no point in the straight line drawn from D to E shall coincide with any point in BD or BE, except the points D and E which were joined.

For because BE makes with BD an angle less than the sum of two right angles, (by Cor. 1 above) it is not in the same straight

line with it. Therefore if the points D and E be joined, there *INTERC12. shall be formed* a three-sided figure or triangle ; and no point in Cor. 7.

the straight line from D to E shall coincide with any point in BD or BE, except the points D and E which were joined. Also the straight line from D to E cannot pass on the side of DC which is remote from E; for then, to arrive at E, it must pass through DC or its prolongation, and two straight lines would inclose a space. Therefore it shall pass on the side on which is E; that is, on which is the angle that is less than the sum of two right angles.

Cor. 4. If any number of straight lines be added one to another in succession, so that at each of the points of junction a straight line drawn to such point makes the adjacent angles together equal to two right angles, all the first-mentioned straight lines shall be in one and the same straight line.

For (by Prop. XIII above) the second is in one straight line with the first ; and in like manner the third with the straight line which is the sum of the second and first; and so on.

Cor. 5. Any straight line within a triangle, as DE, being prolonged shall cut the perimeter.

For if the point D and any point in the perimeter of the triangle,

as F, be joined ; because the angles DFB and + I, 12. DFA aret together equal to two right angles,

DFB is less than two right angles, and (by Cor. 2
above) DF and BF being prolonged will cut one
another. And because the same may be shown
of any other point in the perimeter, if a straight
line of unlimited length DG be turned about D so as to pass
through every point in the perimeter in succession, it will at all

times cut the perimeter; wherefore if it be moved till it also pass Interc.12. through the point E, the straight line DE willf coincide with Cor. 1.

such unlimited straight line to the extent of the length that is common to both, and its prolongation will also be in the same straight line, and consequently will cut the perimeter.

F

D

B

DA

E

с

PROPOSITION XIV.
See Note. PROBLEM.–To bisect a given angle. [That is, to divide it into

two equal angles.]
Let BAC be the given angle. It is required
to bisect it.

First Case; if BAC is less than the sum of two

right angles. Take any point D in AB, and from • 1.3. AC cut off* AE equal to AD. Join DE, and † 1. 1. upon it describet an equilateral triangle DFE on the side remote from A; whereupon may be

B
shown (as in Prop. IX) that the point F shall F
lie between the straight lines DB and EC. Then

join AF; the straight line AF shall bisect the angle BAC. 1 Constr. Because AD ist equal to AE, and AF is common to the two

triangles DAF, EAF, the two sides AD, AF are equal to the • Constr. two sides AE, AF respectively. Also the third side DF is* equal to +1.8.

the third side EF; therefore the angle DAF ist equal to the angle EAF. Wherefore the given angle BAC is bisected by the straight line AF.

Second Case; if the angle BAC is equal to the sum of two right angles, BAC will be in one and the same straight line, and DE be bisected by the point A, and AF be perpendicular to BAC and bisect the angle BAC by dividing it into two right angles.

Third Case ; if the angle BAC is greater than the sum of two right angles (or the angle intended is that formed by the turning of the radius vectus circuitously by the way of G), bisect the

angle between AB and AC which is less than two right angles, *Interc12. by AF as before ; and prolong* FA to G. The angle BAG is Cor. 6.

equal to the angle CAG; or the circuitous angle BAC is

bisected. +1.12. For because the angles BAF, BAG, are togethert equal to

two right angles, and the angles CAF, CAG, are together equal IInterc. 1. to two right angles; BAF and BAG are together equal to CAF •Constr. and CAG. But the angle BAF is* equal to CAF; wherefore the +INTERC. 1. remaining angle BAG ist equal to the remaining angle CAG. Cor. 7.

And by parity of reasoning, in like manner may every other given angle be bisected.

Cor. Only one straight line, can bisect a given angle.

11. 13.

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