a 3. 1.

b conftr. chyp. d 4. I. e hyp. fy. ax. 9.

g hyp. h confir. k 4. 1. 1 hyp. m 16. 1,

a 16. 1.

a 15. 1.

b. 27. 1.

the other fides of the other, each to his correfpondent fide, and the other angle of the one, shall be equal to the other angle of the other.

1. Hypothefis. Let BC be equal to E G, which are the fides that lie between the equal angles, Then I fay BA ED, and AC=DG, and the angle A-EDG. For if it be faid that EDBA, then (a) let EH be made equal to BA, and let the line GH be drawn.

Because AB(b) — HE, and BC (c) EG, and the angle B()E, therefore fhall be the angle EGH (d)= C(e)=DGE. (f)Which is abfurd, therefore AB ED. After the fame manner AC may be proved equal to DG, (a) then will the angle A be equal to EDG.

2. Hyp. Let A B be equal to DE, then I fay BC= EG, and AC DG, and the angle A-EDG. For if EG be greater than B C make EFB C, and join DF. Now becaufe A B (g)=DE, and B C (b) = EF, and the angle B (g) E; therefore will be the angle EFD (k) == C(1)=EGD. (m) Which is abfurd. Therefore is BC EG, and fo as before, AC DG, and the angle ALDG. Which was to be dem.

PRO P. XXVII. Plate I. Fig. 33.

If a right-line EF; falling upon two right-lines AB, CD, makes the alternate angles Ab F, Dc E, equal the one to the other, then are the right-lines AB, CD, parallel.

If AB, CD be faid not to be parallel, produce them till they meet in O, which being fuppofed, the outward angle AF will be (a) greater than the inward angle De E, to which it was equal by Hypothefis. Which things are repugnant.

PRO P. XXVIII. Fig. 33.

If a right-line EF, falling upon two right-lines, AB, CD, makes the outward angle AGE of the one line equal to CH G the inward and oppofite angle of the other on the fame fide, or make the inward angles on the fame fide, AbÍ, СcG, equal to two right-angles, then are the right-lines, AB, CD, parallel.

Hyp. 1. Because by Hypothefis the angle Ab E Cc G, (a) therefore are Bb H, Cc G, the alternate angles equal; And (b) therefore are AB and CD parallel.

Hyp

Hyp. 2. Because by Hypothefis the angle AGH+ CHG to two right-angles,(a)—AGH+BGH,(†) therefore shall the angle CHG-BGH; and (c) therefore AB, CD, are parallel, Which was to he demonfirated.

PRO P. XXIX. Plate I. Fig. 33.

If a right-line EF falls upon two parallels, AB, CD, it will make both the alternate angles DHG, AGH, equal each to the other, and the outward angle BGE equal to the inward and oppofite angle on the fame fide DHG, as alfo the inward angles on the fame fide AGH, CHG, equal to two right-angles:

It is evident, that AGH+CHGtwo right-angles; (a) otherwise AB, CD, would not be parallel, which is contrary to the Hypothefis: But moreover the angle DHG+CHG (6) two right-angles; therefore is DHG (c): = AGH (d) BGE. Which was to be dem:

Coroll. Fig: 7:

Hence it follows that every parallelogram AD having one angle right A, the reft are alfo right.

For A+B (a) two right-angles. Therefore, whereas A is right, (b) B muft be alfo right. By the fame argument are C and D right-angles.

PRO P. XXX. Fig. 21.

Right-lines (ÁB, MN) parallel to one and the fame rightline KL, are also parallel the one to the other.

Let CD cut the three right-lines given any ways. Then because AB, MN are parallel, the angle BII will be (a) =LHI. Alfo becaufe KL and MN are parallel, the angle LHI will be (a) MIC (b) Therefore the angle BEI=MIC; (c) whence AB and MN are parallel. Which was to be demonftrated.

PROP, XXXI. Fig. 34.

From a point given A, to draw a right-line AE, para!lel to a right-line given BC.

From the point A draw a right-line AD to any point of the given right-line; with which at the point thereof (a) A inake an angle DAE-ADC; (b) then will AE and BC be parallel. Which was to be done. PROP.

B

a 31. 1. b 29. I,

C 2 ax.

d 19. ax.
e 13. I.
f 1. ax.

PROP. XXXII. Plate I. Fig. 35.

Of any triangle ABC one fide BC being drawn out, the outward angle ACD shall be equal to the two inward oppofite angles A, B, and the three inward angles of the triangle, A, B, ACB, shall be equal to two right-angles.

From C (a) draw CE parallel to BA. Then is the angle A (b) ACE, and the angle B (b) —ECD. Therefore A+B (c)≈ACE+ECD (d)=ACD. Which was to be demonftrated.

I affirm ACD+ACB (e) two right-angles; (f) therefore A+B+ACB two right-angles. Which was to be demonftrated.

Coroll.

1. The three angles of any triangle taken together are equal to the three angles of any other triangle taken together. From whence it follows,

2. That if in one triangle, two angles (taken feverally, or together) be equal to two angles of another triangle (taken feverally, or together) then is the remaining angle of the one equal to the remaining angle of the other. In like manner, if two triangles have one angle of the one equal to one of the other, then is the fum of the remaining angles of the one triangle equal to the fum of the remaining angles of the other.

3. If one angle in a triangle be right, the other two are equal to a right-angle. Likewife, that angle in a triangle which is equal to the other two, is it self a rightangle.

4. When in an Ifofceles the angle made by the equal fides is right, the other two upon the bafe are each of them half a right-angle.

5. An angle of an equilateral triangle makes two third. parts of a right-angle. For one third of two right-angles is equal to two thirds of one.

Schol.

By the help of this propofition you may know how many right-angles the inward and outward angles of a right-lined figure make; as may appear by thefe two following Theorems.

THE O.

THEOREM I.

All the angles of a right-lined figure do together make wice as many right-angles, abating four, as there are fides of the figure.

From any point within the figure let right-lines be drawn to all the angles of the figure, which shall resolve the figure into as many triangles as there are fides of the figure. Wherefore, whereas every triangle affords two right-angles, all the triangles taken together will make up twice as many right-angles as there are fides. But the angles about the faid point within the figure make up four right; therefore, if from the angles of all the triangles you take away the angles which are about the faid point, the remaining angles, which make up the angles of the figure, will make twice as many right-angles, abating four, as there are fides of the figure. Which was to be demonftrated.

Coroll.

Hence all right-lined figures of the fame fpecies have the fums of their angles equal.

THEOREM II.

All the outward angles of any right-lined figure, taken together, make up four right-angles.

For every inward angle of a figure, with the outward angle of the fame, make two right-angles; therefore all the inward angles, together with all the outward, make twice as many right-angles as there are fides of the figure; but (as has been just fhewn) all the inward angles, with four right, make twice as many right as there as fides of the figure, therefore the outward angles are equal to four right-angles. Which was to be demonfirated.

Coro it.

All right-lined figures, of whatfoever fpecies, have the fums of their outward angles equal.

a 29. 1.

b 4. I.

C 27. I.

a hyp. b

29. I.

C 2. ax.

d 26. 1.

a 27. 1.

PROP. XXXIII. Plate I. Fig. 7.

If two equal and parallel lines AB, CD, be joyned together with two other right-lines, AC, BD, then are thofe lines alfo equal and parallel.

Draw a line from C to B. Now because AB and CD are parallel, and the angle ABC (a) = BCD; and also by hypothefis A B≈CD, and the fide C B common, therefore is AC (b) BD, and the angle ACB (b) = DBC (c) whence alfo AC, BD, are parallel,

PRO P. XXXIV, Fig. 7.

In parallelograms, as ABDC, the oppofite fides AB, CD, and AC, BD, are equal each to the other; and the oppofite angles A, D, and ABD, ACD, are also equal; and the diameter BC bijects the fame.

Because AB, CD, (a) are parallel, (b) therefore is the angle ABC BCD. Alfo because AC, BD, are (a) parallel, (b) therefore is the angle ACB=CBD; (c) therefore the whole angle ACD ABD. After the fame manner is AD. Moreover because the angles ABC, ACB, lie at each end of the fide CB, and are equal to BCD, CBD, (d) therefore is AC BD, and AB‍(d)= CD, and fo the triangle ABC CBD. Which was to be demonfirated.

Schol.

Every four-fided figure ABDC, having the oppofite fides equal, is a parallelogram.

For by 8. 1. the angle ABC BCD; (a) wherefore AB, CD, are parallel. In like manner is the angle BCA

CBD; (a) wherefore AC, BD, are alfo parallel, (b) b 35. def. 1. Therefore ABCD is a parallelogram. Which was to be

Fig. 36.

16.36.

demonftrated.

From hence we may more expeditiously draw a parallel CG to a right-line given, AB, thro' a point affigned, C.

Take in the line AB any point, as E. From the centers E and C at any distance draw two equal circles EF, CD. From the center F with the diftance EC draw a circle FD, which fhall cut the former circle CD in the point D. Then fhall the line drawn CG be parallel to AB, for, as it was before demonstrated, CEFD is a parallelogram.

PROP.

1