If the angles DGA, DGB be equal, (a) then they make two right-angles; if unequal, then from the point G (6) let there be erected a perpendicular GC. Because the angle DGB (c) = to a right-angle +CGD, and the angle DGA (d) = to a right-angle-DGC; therefore the angle DGB-J-DGA (e) = to two right-angles. Which was to be demonstrated.

Corollaries.

1. Hence, if one angle DGB be right, the other DGA is also right; if one acute, the other is obtufe, and so on the contrary.

2. If more right-lines than one stand upon the same right-line at the same point, the angles shall be equal to two right.

3. Two right-lines cutting each other make angles equal to four right-angles.

4. All the angles made about one point make four right-angles; as appears by Coroll. 2.

PROP. XIV. Plate I. Fig. 1.

If to any right-line CG, and a point therein G two rightlines, not drawn from the fame fide, do make the angles CGA, CGB, on each fide equal to two right, the lines, AG, GB, shall make one ftrait-line.

If you deny it, let AG, GF make one right-line; then shall the angle CGA+CGF (a) two right angles (6) = CGA+CGB. Which is (c) abfurd.

PROP. XV. Fig. 21.

If two right-lines AB, CD, cut thro' one another, then are the two angles which are oppofite, viz. CEB, AED, equal one to the other.

For the angle AEC + CEB (a) = to two right rightangles = AEC + AED; (6) therefore CEB = AED. Which was to be demonstrated.

Schol. 1.

If to any right-line AB, and in it a point E, two right lines being drawn CE, ED, and not taken on the fame side, make the vertical (or opposite) angles CEA, and BED equal, those right-lines CE, ED, do meet directly and make one strait line.

a def. 10.

b II. I.

c 19. ax. d 3. ax.

e 2. ax.

a 13. 1. b hyp.

c 9. ax.

a 13. 1. b 3. ax.

For

For two right-angles are (a) equal to the angle CEA+ СЕВ, (6) = СЕB + BED. (c) Therefore CE, ED, are in a strait line. Which was to be demonstrated.

Schol. 2. Plate I. Fig. 21.

If four right-lines EA, EB, EC, ED, proceeding from one point E, make the angles, vertically oppofite, equal the one to the other, each two lines, AЕ, ЕВ, and CE, ED, are placed in one strait line.

For because the angle AEC + AED + CEB + DEB a 4. c 13. 1. (a) = to four right-angles, therefore the angle AEC + b byp. & AED (6) = CEB + DEB = to two right-angles. (c) Therefore CED and AEB are strait-lines. Which was to be demonstrated.

C2. ax.
C 14.1.

a 10. 1. & 1. poft. b3. 1.

c conftr..

₫ 15. 1.

e 4. I.

f 15. 1.

g9. ax.

PROP. XVI. Fig. 22.

One fide, BC, of any triangle ABC being produced, the outward angle ACD will be greater than either of the inward and opposite angles, CAВ, СВА.

Let the right-lines AH, BE, (a) bisect the sides AC, BC; from which lines produc'd, take (6) EF=BE, and HI, (6) = AH, and join FC, and IC; and produce ACG.

Because CE (c) = EA, and EF (c) = EB, and the angle FEC (d) = BEA, the angle ECF (e) shall be equal to EAB. By the like argument is the angle ICH = ABH. Therefore the whole angle ACD (f) (BCG,) (g) is greater than either the angle CAB or ABC. Which was to be demonftrated.

PROP. XVII. Fig. 23.

Two angles of any triangle ABC, which way foever they are taken, are less than two right-angles.

Let the fide BC be produced. Because the angle ACD + ACB (a) = two right-angles, and the angle ACD (6) A, (c) therefore A + ACB than two right-angles. After the same manner is the angle B + ACB than two right-angles. Lastly, the fide AB being produced, the angle A - B will be also less than two right-angles. Which was to be demonstrated.

Coroll.

1. Hence it follows that in every triangle wherein one angle is either right or obtuse, the two others are acute angles.

2. If a right-line AE make unequal angles with another right-line BC, one acute AEB, the other obtufe AEC, a perpendicular AB, let fall from any point A to the other line BC, shall fall on that side the acute angle is of.

For if AC, drawn on the side of the obtufe-angle, be a perpendicular, then in the triangle AEC, shall the angles AEC + ACE be greater than two right-angles. * Which is contrary to the frecedent prop.

3. All the angles of an equilateral triangle, and the two angles of an Isosceles triangle that are upon the base, are acute.

PROP. XVIII. Plate I. Fig. 23.

The greatest fide AC of every triangle ABC fubtends the greatest angle ABC.

From AC (a) take away AF = AB, and join BF. (b) Therefore is the angle AFB = ABF. But AFB (c) C; therefore is ABFC; (d) therefore the whole angle ABC C. After the same manshall be ABC A. Which was to be demonstrated.

PROP. XIX. Fig. 23.

In every triangle ABC, under the greatest angle Bis fubtended the greatest fide AC.

For if AB be supposed equal to AC, then will be the angle B (a) = C, which is contrary to the Hypothefis: And if AB AC, then shall be the angle C (6) B, which is against the Hypothesis. Wherefore rather AC AB; and after the fame manner AC BC. Which was to be demonstrated.

PROP. XX. Fig. 24.

Of every triangle ABC, tawo fides BA, AC, any way taken, are greater than the fide that remains BC.

Pro

Produce the line BA, (a) and take AD-AC, and draw the line DC; (6) then shall the angle D be equal to ACD; (c) therefore is the whole angle BCDED; (d) therefore BD (e) (BA+AC) BC. Which was to be demonstrated.

PROP. XXI. Plate I. Fig. 25.

1

If from the utmost points of one fide BC, of a triangle ABC, two right-lines BD, CD, be drawn to any point within the triangle, then are both those two lines shorter than the two other fides of the triangle BA, CA; but contain a greater angle, BDC.

Let BD be produced to E. Then is CEED (a) CD, and BD common to both, (6) then shall be DB+ DE-FECCD+BD. Again, BA+AE (a) [ BE; (6) therefore BA+AC-BE-EC. Wherefore 1. BA+ AC-BD+DC. 2. The angle BDC (c) DEC (c) A. Therefore the angle BDC-A. Which was to be demonstrated.

PROP. XXII. Fig. 26.

To make a triangle FKG of three right-lines FK, FG, GK, which shall be equal to three right-lines given A,B,C. Of which it is necessary that any two taken together be longer than the third.

From the infinite line DE (a) take DF, FG, GH, equal to the lines given A,B,C. Then if from the (6) centers F and G at the distances of FD and GH, two circles be drawn cutting each other in K, and the right-lines KF, KG be joined, the triangle FKG shall be made, (c) whose fides FK, FG, GK, are equal to the three lines DF, FG, GH, (d) that is, to the three lines given A, B, C. Which was to be done.

PROP. XXIII. Fig. 27, and 28.

At a point A in a right-line given AH, to make a rightlined angle A equal to a right-lined angle given D.

(a) Draw the right-line CF cutting the fides of the angle given any ways; (6) make AG CD; upon AG (c) raise a triangle equilateral to the former CDF, fo that AH be equal to DF, and GH to CF. then shall you have the angle A (d) = D. Which was to be done.

PROP. XXIV. Plate I. Fig. 29, 30.

If tavo triangles ABC, DEF have two fides of the one triangle AB, AC, equal to two fides of the other triangle DE, DF, each to each, and have the angle A greater than the angle EDF contained under the equal right-lines, they shall have also the Base BC greater than the base EF.

(a) Let the angle EDG be made equal to A, and the fide DG (6)=DF (c) =AC; and let EG, and FG be joined.

1. Cafe. If EG falls above EF; Because AB (d) =DE, and AC (e) DG, and the angle A (e) EDG; (f) therefore is BC-EG. But because DF (e) =DG; (g) therefore is the angle DFG=DGF; (b) therefore is the angle DFG EGF, and by consequence the angle EFG, (b) EGF; (k) wherefore EG (BC) EF. 2. Cafe. If the base EF coincides with the base EG, (1) it is evident that EG (BC) EF. Fig. 29, 31. 3. Cafe. If EG falls below EF, then because DG+ GE (m) DF-FE, if from both be taken away DG, DF which are equal; EG (BC) remains (n) Which was to be demonstrated. Fig. 29, 32.

PROP. XXV. Fig. 29, 30.

EF.

If two triangles ABC, DEF, have two fides AB, AC, equal to two fides DE, DF, each to each, and have the bafe BC greater than the base EF, they shall also have the angle A contained under the equal right-lines greater than the angle EDF.

a 23. 1. b 3. 1. c hyp. d byp.

e constr. f 4. I. g 5. 1. h 9. ax. k 19. 1.

1 9. ax.

m 21. 1. n 5. ax.

a 4. 1.

For if the angle A be faid to be equal to EDF, (a) then is the base BC=EF, which is against the Hypothesis. If it be faid the angle A EDF, then (6) b 24.1 will be BCEF, which is also against the Hypothesis. Therefore ACEDF. Which was to be demonstrated.

PROP. XXVI. Fig. 29, 31.

If two triangles BAC, EDG, have two angles of the one B, C, equal to two angles of the other E, DGE, each to his correspondent angle, and have also one side of the one equal to one fide of the other, either that fide which lyeth betwixt the equal angles, or that which is fubtended under one of the equal angles; the other fides also of the one shall be equal to the