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lines meeting one another towards neither parts are parallel: there- Book I, fore AB is parallel to CD.

Wherefore if a straight line falling upon two straight lines make the alternate angles equal to one another, the straight lines will be parallel. Which was to be demonstrated.

PROP. XXVIII.

If a straight line falling upon two ftraight lines make the outward angle equal to the inward and oppofite, and towards the fame parts; or make the inward angles and towards the fame parts. equal to two right angles: the straight lines will be parallel to one another.

For let the straight line EF falling upon the two straight lines AB, CD make the outward angle EGB equal to the inward and oppofite and towards the fame parts the angle GHD; or the inward and towards the fame parts, the angles BGH, GHD equal to two right angles; I say that AB is parallel to CD.

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Again because the angles BGH, GHD are (by fupp.) equal to two right angles; and also the angles AGH, BGH are equal (by prop. 13.) to two right angles; therefore (by com. not. 1.) the angles AGH, BGH are equal to the angles BGH, GHD: let the common angle BGH be taken away, therefore (by com. not. 3.) the remaining angle AGH is equal to the remaining angle GHD: and they are alternate; wherefore (by prop. 27.) AB is parallel to

CD.

Wherefore if a ftraight line falling upon two ftraight lines make the outward angle equal to the inward and oppofite and towards the fame parts; or make the inward angles and towards the fame parts

equal

Book I. equal to two right angles: the ftraight lines will be parallel to one another. Which was to be demonstrated.

PROP. XXIX.

Or a straight line falling upon the two parallel straight lines makes the alternate anglès equal to one another; and the outward angle equal to the inward and oppofite and towards the fame parts; and the inward angles and towards the fame parts equal to two right angles.

A

E

G

B

For let the straight line EF fall upon parallel straight lines, the ftraight lines AB, CD: I fay that it makes the alternate angles AGH, GHD equal; and the outward angle EGB equal to GHD the inward and oppofite and towards the fame parts; and BGH, GHD the inward and towards the fame parts equal to two right angles. For if AGH be unequal to GHD, one of them is greater: let AGH be the greater and because AGH is greater than GHD, let BGH which is common be added; therefore (by com. not. 4.) the angles AGH, BGH are greater than BGH, GHD: But also the angles AGH, BGH are (by prop. 13.) equal to two right angles; and therefore the angles BGH, GHD are less than two right angles; but those ftraight lines which are produced indefinitely from angles less than two right angles meet one another (by com. not. 11.); therefore AB and CD being produced indefinitely will meet one another; but they do not meet, because they are supposed parallel: therefore AGH is not unequal to GHD, therefore it is equal.

H

D

F

But the angle AGH is equal to the angle EGB (by prop. 15.); and therefore (by com. not. 1.) EGB is equal to GHD.

Let the common angle BGH be added: wherefore the angles EGB, BGH are equal to the angles BGH, GHD: but the angles EGB, BGH are (by prop. 13.) equal to two right angles; and therefore the angles BGH, GHD are (by com. not. 1.) equal to two right angles.

Or a straight line, therefore, falling upon the two parallel ftraight lines, makes the alternate angles equal to one another; and

the

the outward angle equal to the inward and oppofite and towards the Book I. fame parts; and the inward angles and towards the fame parts equal to two right angles. Which was to be demonstrated.

PRO P. XXX.

Straight lines parallel to the fame ftraight line are also parallel to one another.

Let each of the straight lines AB, CD be parallel to EF: I fay also that AB is parallel to CD.

A

For let the ftraight line GK fall upon them. And because the ftraight line GK hath fallen upon parallel straight lines AB, EF therefore (by prop. 29.) the angle AGH is equal to GHF: again, be- E cause the straight line GK hath fallen upon the parallel ftraight lines EF, CD C the angle GHF is equal to GKD: but

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the angle AGK hath been proved equal to GHF: and therefore (by com. not. 1.) AGK is equal to GKD; and they are alternate angles: therefore AB is parallel to CD (by prop. 27.)

Wherefore the ftraight lines which are parallel to the fame ftraight line, are parallel to one another. Which was to be demonftrated.

PROP. XXXI.

Through a given point to draw a straight line parallel to a given ftraight line

Let the given point be the point A; and the given straight line. the ftraight line BC; it is required through the point A to draw a ftraight line parallel to the ftraight line BC.

E

A F

Let any point whatever the point D be taken in BC; and let AD be joined. And let the angle DAE be made, with the straight line AD and at the point A in it, equal to the angle ADC (by prop. 23.): and let the straight line AF be produced in a straight line with AE.

B

D

C

And

Book I.

And because the straight line AD falling upon the two straight lines BC, EF hath made the alternate angles, the angles EAD, ADC equal to one another, therefore (by prop. 27.) EF is parallel to BC.

Wherefore through a given point the point A a straight line EAF hath been drawn parallel to the given straight line BC. Which was to be done.

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One of the fides of any triangle being produced, the outward angle is equal to the two inward and opposite angles: and the three inward angles of the triangle are equal to two right angles.

Let there be a triangle, the triangle ABC, and let one of its fides BC be produced to the point D: I say that the outward angle ACD is equal to the two inward and oppofite ones CAB, ABC; and that the three inward angles of the triangle viz. ABC, BCA, CAB are equal to two right angles.

For let CE be drawn (by prop. 31.) through the point C parallel to the straight line AB.

A

E

And because AB is parallel to CE and AC hath fallen upon them; the alternate angles BAC, ACE are equal (by prop. 27.): Again, because AB is parallel to CE and the straight line BD hath fallen upon them; the outward angle ECD is equal to the inward and oppofite ABC (by prop. 29.): but ACE has been alfo fhewn to be equal to BAC; the whole therefore, the outward angle ACD is equal to the two inward and and oppofite ones BAC, ABC.

B

C D

Let the common angle ACB be added therefore the angles ACD, ACB are equal to the three angles ABC, BAC, BCA; but ACD, ACB are (by prop. 13.) equal to two right angles; and therefore (by com. not. 1.) ACB, CBA, CAB are equal to two right angles.

Wherefore one of the fides of any triangle being produced, the outward angle is equal to the two inward and oppofite angles: and the three inward angles of the triangle are equal to two right angles. Which was to be demonstrated.

PROP.

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The straight lines joining equal and parallel ftraight lines towards the fame parts, are also themselves equal and parallel.

Let the ftraight lines AB, CD be equal and parallel; and let the ftraight lines AC, BD join them towards the fame parts: I fay that AC, BD are alfo equal and parallel.

For let BC be joined.

:

D

B

C

A

And because AB is parallel to CD, and BC hath fallen upon them; the alternate angles, the angles ABC, BCD are (by prop. 29.) equal to one another and becaufe (by fupp.) AB is equal to CD, and the ftraight line BC common; certainly the two AB, BC are equal to the two DC, CB; and the angle ABC is equal to BCD; therefore (by prop. 4.) the base AC is equal to the base BD; and the triangle ABC is equal to the triangle BCD; and the remaining angles will be equal to the remaining angles, under which the equal fides are extended, each to each; therefore the angle ACB is equal to the angle CBD: and because the straight line BC falling upon the two straight lines AC, BD hath made the alternate angles ACB, CBD equal to one another; therefore (by prop. 27.) AC is parallel to BD; and it has been demonftrated also to be equal.

Wherefore the straight lines joining equal and parallel straight lines towards the same parts are also themselves equal and parallel. Which was to be demonstrated.

Book I.

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The oppofite fides and alfo the oppofite angles of parallelogram fpaces are equal to one another, and the diameter cuts the spaces in halves.

Let there be a parallelogram, the parallelogram ACDB and its diameter the straight line BC: I say that the oppofite fides and alfo the angles of the parallelogram ACDB are equal to one another, and the diameter BC cuts it in halves.

VOL. I.

E

For

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