a 22. 3.

b byp.

C 3. ax.

d 21. 1.

Corol.

r. Hence, if one fide AB of a quadrilateral, defcribed in a circle, be produced, the external angle EBC is equal to the internal angle ADC, which is oppofite to that ABC, and adjacent to EBC, as appears by 13. 1. and 3. ax.

2. A circle cannot be described about a Rhombus because its opposite angles are greater, or less than two right-angles.

Schol. Plate II. Fig. 24.

If in a quadrilateral ABCD, the angles A, and C, which are oppofite, be equal to two right-angles, then a circle may be described about that quadrilateral.

For a circle will pass through any three angles B,C,D, (as appears by 5.4.) I say that it shall also pass thro' A, the 4th angle of such a quadrilateral: For if you deny it, let the circle pass thro' F: Therefore the rightlines BF, FD, BD being drawn, the angle C+F (a)= 2 right (b)=C+A; wherefore A (c) is equal to F, (d) Which is abfurd.

PROP. XXIII. Fig. 25.

Two like and unequal fegments of circles ABC, ADC, cannot be set on the fame right-line AC, and on the same fide thereof.

For if they are said to be like, draw the line CB cutting the circumference in D and B, join AB and AD. a 10. def. 3. Because the segments are supposed like, (a) therefore is the angle ADC=ABC. (b) Which is abfurd.

b 16. 1.

a 23.3.

b 10. 3. c 8. ax.

PROP. XXIV. Fig. 26.

Like segments, of circles ABC, DEF, upon equal right-lines AC, DF, are equal one to the other. The base AC being laid on the base DF, will agree with it, because AC= DF. Therefore the segment ABC shall agree with the fegment DEF (for otherwise it shall fall either within or without) and if so (a) then the segments are not like, which is contrary to the Hypothesis, and at least it shall fall partly within and partly without, and so cut in three points, (6) which is absurd. (e) Therefore the segment ABC=DEF. Which was to be demonstrated.

PROP. XXV. Plate II. Fig. 27.

A segment of a circle ABC, being given, to describe the whole circle whereof that is a segment.

Let two right-lines be drawn AB, BC, which bisect in the points D and E. From D and E draw the perpendiculars DF, EF, meeting in the point F. I say this point shall be the center of the circle. For the center shall be as well in (a) DF as EF, therefore it must be in the point F, which is common to them both. Which was to be done.

a cor. 1. 3.

PROP. XXVI. Fig. 28, 29.

In equal circles GABC, HDEF, equal angles stand upon equal parts of the circumference, AC, DF; whether those angles be made at the centers G, H, or at the circumferences, B, E.

Because the circles are equal, therefore is GA=HD, and GC=HF; also by Hypothesis the angle G=H, a 4.1. (a) therefore AC=DF. Moreover the angle B (6) Gb 20.3. (c) = H (b) =E. (d) Therefore the segments ABC, DEF are like, and (e) confequently equal; (f) whence the remaining segments also AC, DF, are equal. Which was to be demonstrated.

Schol. Fig. 30.

In a circle ABCD, let an arch AB be equal to DC; then shall AD be parallel to BC. For the right-line AC being drawn, the angle ACB (a) = CAD; wherefore by 27. 1. the faid fides are parallel.

PROP. XXVII. Fig. 28, 29.

In equal circles GABC, HDEF, the angles ftanding upon equal parts of the circumference. AC, DF, are equal between themselves, whether they be made at the centers G, H, or at the circumferences, B, E.

For if it be possible, let one of the angles AGC be DHF, and make AGI=DHF; thence is the arch Al (a) =DF (b) = AC. (c) Which is abfurd,

Schol. Fig. 31.

A right-line EF, which being drawn from A the middle point of any periphery BC, touching the circle, is parallel

to

c hyp. dio.def.3.

e 24.3. f 3. ex.

a 26. 3.

a 26. 3. b hyp.1 c 9. ax.

a 27.3. b byp.

C 4. 1.

to the right-line BC, fubtending the said periphery,

From the center D draw a right-line DA to the point of contact A, and join DB, DC.

The fide GD is common, and DB=DC, and the angle BDA (a) =CDA, (because the arches BA, CA are (b) equal) therefore the angles at the base DGB, DGC are (c) equal, and (d) consequently right; but the inward

d 10. def. 1. angles GAE, GAF are also (e) right; (f) therefore BC, e hyp. f 28. 1.

a hyp.. b 8. 1.

с 26. 3.

d 3. ax.

a hyp. b. 27. 3. C 4.1.

a conftr.

b 12, ax.

C 4. I.

EF are parallel. Which was to be demonstrated.

PROP. XXVIII. Plate IL. Fig. 28, 29.

In equal circles GABC, HDEF, equal right-lines AC, DF, cut off equal parts of the circumference, the greatest ual to the greatest DEF, and the leaft ALC to the

leaft DKF.

From the centers G, H, draw GA, GC, and HD, HF. Because GA=HD, and GC=HF, and AC (a)=DF, (b) therefore is the angle G=H; (c) whence the arch ALC=DKF; (d) and so the remaining arch ABC= DEF. Which was to be demonstrated.

But if the subtended line AC be or than DF, then in like manner will be the arch AC be

than DF.

PROP. XXIX. Fig. 28, 29.

or

In equal circles GABC, HDEF, the right-lines AC, DF, which subtend equal peripheries ABC, DEF, are equal. Draw the lines GA, GC, and HD, HF. Because GA=HD, and GC=HF, and (because the arches AC, DF are (a) equal) the angle G (6) = H, (c) therefore is the base AC=DF. Which was to be demonstrated. This and the three precedent propofitions may be understood alfo of the fame circle.

PROP. XXX. Fig. 32.

To cut a Periphery given ABC into two equal parts. Draw the right-line AC, and bisect it in D; from D draw a perpendicular DB meeting with the arch in B, it shall bisect the same.

For join AB, and CB. The fide DR is common, and AD (a) = DC, and the angle ADB (b) = CDB. (c) Therefore AB = BC; (4) whence the arch AB=BC. Which was to be done.

PROP. XXXI. Plate II. Fig. 33.

In a circle the angle ABC, which is in the femicircle, is a right-angle; but the angle, which is in the greater Segment BAC, is less than a right-angle, and the angle which is in the leffer segment BFC is greater than a rightangle. Moreover, the angle of the greater segment is greater than a right-angle, and the angle of the lefsfer segment is less than a right-angle.

a 5. 1.

b 2. ax.
C 32. 1.
d 10.def.1.
ecor.17.1.
f 22. 3.

From the center D draw DB. Because DB = DA, therefore is the angle A (a) = DBA, and the angle DCB (a) = DBC, (b) therefore the angle ABC=A+ ACB, (c) = EBC, (d) fo that ABC and EBC are rightangles. W. W'. D. (e) Therefore BAC is an acute-angle. W.W. D. And further, whereas BAC + BFC (f) = 2 right-angles, therefore BFC is an obtufe-angle. Lastly, the angle contained under the right-line CB, and the arch BAC is greater than the right-angle ABC; but the angle made by the right-line CB, and the periphery of the leffer segment BFC (g) is less than the 59. ax. right-angle EBC. Which was to be demonstrated.

Schol.

In a right-angled triangle ABC, if the hypothenuse (or line fubtending the right-angle) AC be bisected in D, a circle drawn from the center D, through the point A, shall also pass through the point B; as you may easily demonftrate from this prop. and 21. 1.

PROP. XXXII. Fig. 34.

If a right-line AB touch a circle, and from the point of contact be drawn a right-line CE, cutting the circle, the angles ECB, ECA, which it makes with the tangent line, are equal to those angles EDC, EFC, which are made in the alternate segments of the circle.

Let CD, the fide of the angle EDC be perpendicular to AB, (a) for it is to the same purpose, (b) therefore CD is the diameter, (c) therefore the angle CED in a semicircle is a right-angle, (d and therefore the angle D+DCE = to a right-angle (c) = ECB + DCE. (f) Therefore the angle D = ECB, Which was to be dem. Now whereas the angle ECB + FCA (g) = 2 rightangles (6) = D + F, from both of these take away ECB and D, which are equal, (4) there remains ECA F. Which was to be dem.

PROP.

a 26. 3. b 19. 3. C 31.3. d 32. 1.

e conftr. f 3. ax. g 13. 1.

h 22. 3.

k 3. ax.

a 23. I.

c conftr.

с 6. 1.

PROP. XXXIII. Plate II. Fig. 35.

Upon a right-line AB to describe a segment of a circle AIEB which shall contain an angle AIB, equal to a rightlined angle given C.

(a) Make the angle BAD=C. Through the point A draw the line AE perpendicular to HD. At the other end of the line given AB make an angle ABF=BAF, one of the fides of which shall cut the line AE in F; from the center F, through the point A, describe a circle, which shall pass through B. (Because the angle FBA (6) FAB, and (c) therefore FB=FA) AIB is the seginent fought. For because HD is perpendicular

d cor. 16. 3. to the diameter AE, therefore HD (d) touches the circle

a 32. 3. f constr.

a 17.3. b 23. 1.

C 32.3.

d constr.

4

which AB cuts, And therefore the angle AIB (e) BAD (f) = C. Which was to be done.

PROP. XXXIV. Fig. 36.

From a circle given ABC to cut off a segment ABC containing an angle B, equal to a right-lined angle given D.

(a) Draw a right-line EF which shall touch the circle given in A, (b) let AC be drawn making an angle FAC =D. Also draw (6) AB, making the angle BAE=D, and join B, C. The line AB shall cut off the segment ABC containing an angle B (c) =CAF (d)=D. Which was to be done.

PROP. XXXV. Fig. 37.

If in a circle DBCA two right-lines AB, DC cut each other, the rectangle comprehended under the segments AE, EB, of the one, shall be equal to the rectangle comprehended under the segments CE, ED of the other.

1. Cafe. If the right-lines cut one the other in the center, the thing is evident.

2. Cafe. If one lineAB (Fig. 38.) passes thro' the center F, and bisects the other line CD, then draw FD. Now

a 5. 2. b sch. 48. 1. the rectangle AEB+FEq (a) =FBq. (6) =FDq. (c) = EDq+FEq (d)=CED+FEq. (e) Therefore the rectangle AEB=CED. Which was to be demonstrated.

C 47. I.

d hyp.

e 3. ax.

3. Cafe If one of the lines AB (Fig. 39.) be the diameter, and cut the other line CD unequally, bisect CD by FG, a perpendicular from the center.