a 3. 1.

b constr.

chyp.

d 4. I.

e hyp.

f y. ax.

g hyp.

h confir.

k 4. 1.

1 hyp.

m16.1.

a 16. 1.

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

1. Hypothesis. Let BC be equal to EG, 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 EDE BA, 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 shall 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, (d) then will the angle A be equal to EDG.

2. Hyp. Let AB be equal to DE, then I say BC= EG, and AC = DG, and the angle A-EDG. For if EG be greater than BC make EF = BC, and join DF. Now because 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 so as before, AC = DG, and the angle A = LDG. Which was to be dem.

PROP. XXVII. Plate I. Fig. 33.

If a right-line EF; falling upon two right-lines AB, CD, makes the alternate angles AbF, DE, equal the one to the other, then are the right-lines AB, CD, parallel. If AB, CD be said not to be parallel, produce them allel till they meet in O, which being supposed, the outward angle AbF will be (a) greater than the inward angle DE, to which it was equal by Hypothefis. Which things are repugnant.

PROP. 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 CHG the inward and opposite angle of the on the same fide, or make the inward angles on the same fide, AbH, CcG, equal to two right-angles, then are the right-lines, AB, CD, parallel.

other

Hyp. 1. Because by Hypothesis the angle AbE =CcG, (a) therefore are BbH, CG, the alternate angles equal; And (b) therefore are AB and CD paral

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

PROP. 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 same fide DHG, as alfo the inward angles on the same fide AGH, CHG, equal to two right-angles:

It is evident, that AGHCHG two right-angles; (a) otherwise AB, CD, would not be parallel, which is contrary to the Hypothesis: 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

right

one angle right A, the rest are also right.

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

PROP. XXX. Fig. 21.

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

are

Let CD cut the three right-lines given any ways. Then because AB, MN are parallel, the angle BI will be (a) =LHI. Also because KL and MN 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 demonstrated.

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 make an angle DAE=ADC; (b) then will AE and BC be parallel. Which was to be done.

B

PROP.

a 13. 1. b 3. ax. C 27. 1.

a 13. ax. b 13. 1.

c 3 ax.

d 15. I.

2 29. 1. b 3. ax.

a 29.
a 29. 1.
bi. ax.
C 27. 1.

a 23.12 b 27. 1.

a 31. 1. b 29. 1.

C 2 aх.

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 demonstrated.

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

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 severally, 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. Likewise, that angle in a triangle which is equal to the other two, is it self a rightangle.

4. When in an Isosceles the angle made by the equal sides is right, the other two upon the base 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 proposition you may know how many right-angles the inward and outward angles of a right-lined figure make; as may appear by these two following Theorems.

THEO.

!

THEOREM I.

All the angles of a right-lined figure do together make twice 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 refolve 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 said 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 said 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 same species 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 shewn) 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 demonftrated.

Coroll.

All right-lined figures, of whatsoever species, have the sums of their outward angles equal.

}

1

20

2 29. 1.

b 4. I. C 27. I.

a hyp.
b 29. 1.

C 2. ax.

d 26. 1.

a 27. 1.

b 35. def. 1.

Fig. 36.

The first Book of

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 those lines also 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 ABCD, and the fide C B common, therefore is AC (b) = BD, and the angle ACB (b) DBC (c) whence alfo AC, BD, are parallel,

PROP. XXXIV. Fig. 7.

In parallelograms, as ABDC, the opposite 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 bisects the same.

Because AB, CD, (a) are parallel, (b) therefore is the angle ABC=BCD. Also because AC, BD, are (a) parallel, (6) therefore is the angle ACB=CBD; (c) therefore the whole angle ACD = ABD. After the fame manner is A = D. Moreover because the angles ABC, ACB, lie at each end of the side CB, and are equal to BCD, CBD, (d) therefore is AC = BD, and AB (d) = CD, and so the triangle ABC = CBD. Which was to te demonfirated.

Schol.

Every four-fided figure ABDC, having the opposite 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 also parallel. (6) Therefore ABCD is a parallelogram. Which was to be demonftrared.

From hence we may more expeditioussy draw a parallel CG to a right-line given, AB, thro' a point assigned, 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 distance EC draw a circle FD, which shall cut the former circle CD in the point D. Then shall the line drawn.CG be parallel to AB, for, as it was before demonftrated, CEFD is a parallelogram.

PROP.

1