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BFG: But [by the fuppofition] it is likewife equal to it; which is impoffible: Therefore A B, C D continued out towards BD will not meet. After the same manner we demonftrate, that these lines will not meet towards a c: but right lines, which being either way produced, and do not meet, are [by def. 35.] parallel to one another. Therefore A B is parallel to CD,

Wherefore if a right line falling upon two right lines makes the alternate angles equal to one another, these two right lines fhall be parallel. Which was to be demonstrated.

PROP. XXVIII. THEOR.

If a right line falling upon two right lines makes the outward angle of the one equal to the inward oppofite angle of the other, on the fame fide; or the inward angles on the fame fide together equal to two right angles: thefe right lines fhall be parallel to one another,

For let the right line EF, falling upon the two right lines AB, CD, make the outward angle E G B equal to the inward oppofite angle G HD, on the fame fide; or the inward angles B G H, GHD, on the fame fide together equal to two right angles; I fay, the right line A B is parallel to the right line C D.

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[by prop. 27. 1.] A B is parallel to C D.

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Secondly, Because the angles B G H, GHD are equal to two right angles, and [by prop. 13.] the angles AGH, BGH alfo equal to two right angles; the angles AGH, BGH will be equal to the angles BGH, G H D. Take away the common angle BGH, and then the remaining angle AGH is equal to the remaining angle G H D ; and they are alternate angles: Therefore A B [by prop. 27.] is parallel to c D.

D

If

If therefore a right line falling upon two right lines makes the outward angle of the one equal to the inward oppofite angle of the other, on the fame fide; or the inward angles, on the fame fide, together equal to two right angles; these right lines fhall be parallel to one another. Which was to be demonftrated.

PROP. XXIX. THEOR.

If a right line falls upon two parallel right lines, it makes the alternate angles equal to one another; the outward angle equal to the inward and oppofite angle on the fame fide; and the inward angles on the fame fide together equal to two right angles.

For let the right line E F fall upon the parallel right lines A B, CD: Ifay, it makes the alternate angles A G H, GHD equal to one another; the outward angle E GB equal to the inward oppofite angle G HD on the fame fide; and the two inward angles B GH, GHD on the fame fide together equal to two right angles.

For if the angle AGH be unequal to the angle G H D, one of them is the greater, which let be AG H. Then because the angle AGH is greater than the angle G H D, HD, put the common angle B G H to both. Therefore will the angles AGH, BGH be greater than the angles BGH, GHD.

E

G

A

B

D

C

H

F

But the angles AGH, BGH are [by prop. 13.] equal to two right angles: Therefore the angles BG H, GHD are lefs than two right angles. But [by ax. 11.] right lines infinitely produced from angles lefs than two right angles, will meet: Therefore the right lines A B, C D infinitely produced, will meet one another; but they do not meet, fince they are supposed to be parallel. Therefore the angle AGH is not unequal to the angle GHD. It is therefore equal to it.

Secondly, The angle A G H is equal [by prop. 15.] to the angle EGB; therefore alfo EGB will be equal to

G H D.

Thirdly,

Thirdly, Put BGH, which is common, to both; then the angles EG B, B G H are equal to the angles BGH, GH D. But E G B, B G H together, are [by prop. 13.] equal to two right angles: Therefore alfo B G H, G H D are, together, equal to two right angles.

Therefore if a right line falls upon two parallel right lines, it will make the alternate angles equal; the outward angle equal to the inward and oppofite angle on the fame fide; and the inward angles, on the fame fide, together, equal to two right angles. Which was to be demonftrated.

i Scarborough, in his annotations upon this propofition, towards the end, has advanced a ftrange paradox, viz. that two right lines drawn from angles lefs than two right angles, may in fome manner be for ever prolonged, yet hall they never meet together; and (as he thinks) has given a demonftration thereof. But he has miftook the matter. Let us fee what he fays.

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Let two right lines A B, CD be cut, by AC, making the inward angles BAC, DCA less than two right angles. Now let AC be cut into halves, or otherwife, in E: equal to E A let be put A F, and to EC, CG; then draw FG. Again, let FG be cut in half in H, and equal to HF let be put F B, and to HG, GD; then draw BD: I fay, that C

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the lines A B, C D may for ever be thus prolonged, yet never fhall they meet together. For, if poffible, let them meet in the point L; therefore B D being cut in half in K the line K в shall be equal to в L, and K D to DL. Wherefore of the triangle B D L, the fides D L, LB fhall be equal to the third fide B D ; which is impoffible [by 20. 1.]. Therefore the lines A B, CD, drawn from angles lefs than two right, may for ever be prolonged and never meet together. Which was to be demonftrated.Thefe are Scarborough's own words. Now, I fay, he is deceived here; for his two right lines A B, CD will really meet in L, neither a L nor c L being infinite; and his ultimate tri angle DBL will become infinitely fmall, the points D, B, L all coinciding together; fo that here the 20th propofition becomes unapplicable. And I think, from all that he fays, we can obtain no more than this, viz. that an infinite number of right lines FG, BD, all fhorter and fhorter, and nearer and nearer approaching, may be drawn within the triangle ACL; and all the prolongations that a B, CD will acquire, by an infinite

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number

number of particular operations of his, will be only the finite lengths BL, D L.

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Thofe right lines that are parallel to the fame right line, are parallel to one another.

Let each of the right lines A B, C D be parallel to the right line EF: I fay alfo, A B will be parallel to C D. For let the right line G K fall upon them.

Then because the right line G K falls upon the parallel

right lines A B, E F, the angle AGH [by prop. 29.] is equal to the angle GHF.

G

B

A

H

F

Again, because the right line

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D

GK falls upon the parallel right lines E F, C D, the angle GHF is [by prop. 29.]

equal to the angle G K D. But it has been proved, that the angle AG K is alfo equal to the angle GHF; and therefore the angle A G K will be equal to the angle G K D ; and they are alternate angles: Therefore [by prop. 27.] the right line A B is parallel to C D.

Wherefore thofe right lines that are parallel to the fame Which was to be right line, are parallel to one another. demonftrated.

PROP. XXXI. PROBL.

To draw a right line through a given point, parallel to a given right line.

Let the given point be A, and the given right line be BC: It is required to draw a right line through the given point A, parallel to the given right line B C.

E

A

F

Take any point D in в C, and join A D; and [by prop. 23.] with the right line AD, at the point A therein, make the angle D A E equal to the angle A D c, and continue out the right line E A to F.

B

D

Then

Then becaufe the right line A D falling upon two right lines EF, BC makes the alternate angles E AD, AD C equal; the right line E F will be parallel to the right line BC [by prop. 27.].

Therefore the right line E AF is drawn through the given point A, parallel to the given right line B C. Which was to be done.

PROP. XXXII.

THEOR.

If one fide of any triangle be produced, the outward angle is equal to the inward oppofite angles taken together; and the three inward angles of every triangle altogether are equal to two right angles *.

Let A B C be a triangle, and one of its fides B C produced to D: I fay, the outward angle A CD is equal to the two inward oppofite angles c A B, AB C, taken together; and the three inward angles A B C, B C A, C A B altogether are equal to two right angles.

For [by prop. 31.] draw the right line CE through the point c parallel to the right line A B.

Then because A B is parallel

B

A

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E

D

to c E, and the right line A C falls upon them; the alternate angles B A C, ACE [by prop. 29.] are equal to one another. Again, because A B is parallel to c E, and the right line B D falls upon them; the outward angle E CD is equal to the inward oppofite angle A B C. But it has been proved, that the angle A C E is equal to the angle B A C; wherefore the whole outward angle ACD is equal to both the inward oppofite angles B AC, A B C.

Let the angle AC B, which is common, be added to both; then the angles A C D, A C B, together, are equal to all the three angles A B C, B C A, C AB. But [by prop. 13.] both the angles A c D, A C B are equal to two right angles: Therefore all the angles АСВ, СВА, САВ are equal to two right angles.

Therefore if one side of any triangle be produced, the outward angle is equal to both the inward oppofite angles;

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and

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