sary, to E, and join EC: and because the angle ABE is equal to the angle BAE, the straight line BE is equal (6. 1.) to EA: and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each; and the angle ADE is equal to the angle CDE, for each of them is a right angle; therefore the base AE is equal (4. 1.) to the base EC: but AE was shown to be equal to EB, wherefore also BE is equal to EC: and the three straight lines AE, EB, EC are therefore equal to one another; wherefore (9. 3.) E is the centre of the circle. From the centre E, at the distance of any of the three AE, EB, EC, describe a circle, this shall pass through the other points; and the circle of which ABC is a segment is described: also, it is evident, that if the angle ABD be greater than the angle BAD, the centre E falls without the segment ABC, which therefore is less than a semicircle; but if the angle ABD be less than BAD, the centre E falls within the segment ABC, which is therefore greater than a semicircle: Wherefore, a segment of a circle being given, the circle is described of which it ís a segment. Which was to be done.

PROP. XXVI. THEOR.

In equal circles, equal angles stand upon equal arches, whether they be at the centres or circumferences.

Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences: the arch BKC is equal to the arch ELF.

Join BC, EF; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two

sides BG, GC, are equal to the two EH, HF; and the angle at & is equ l to the angle at H; therefore the base BC is equal (4. 1.) to the base EF: and because the angle at A is equal to the angle at D, the segment BAC is similar (9. def. 3.) to the segment EDF; and they are upon equal straight lines BC, EF; but similar segments of circles upon equal straight lines are equal (24. 3.) to one another, therefore the segment BAC is equal to the segment EDF: but the whole circle ABC is equal to the whole DEF; therefore the remaining segment BKC is equal to the remaining segment ELF, and the arch BKC to the arch ELF. Wherefore in equal circles, &c. Q. E. D.

+ PROP. XXVII. THEOR.

In equal circles, the angles which stand upon equal arches are equal to one another, whether they be at the centres or circumferences.

Let the angles BGC, EHF at the centres, and BAC, EDF at the circumferences of the equal circles ABC, DEF stand upon the equal arches BC, EF: the angle BGC is equal to the angle EHF, and the angle BAC to the angle EDF.

If the angle BGC be equal to the angle EHF, it is manifest (20. 3.) that the angle BAC is also equal to EDF. But, if not, one of them is the greater: let BGC be the greater, and at the point G, in the straight line BG, make the angle (23. 1.) BGK equal to the angle EHF And because equal angles stand upon equal arches (26. 3.), when they are at the centre, the arch BK is equal to the arch EF: but EF is equal to C; therefore also BK is equal to BC, the less to the greater, which is impossible. Therefore the angle BGC is not unequal to the angle EHF; that is, it is equal to it: and the angle at A is half of the

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angle BGC, and the angle at D half the angle EHF: therefore the angle at A is equal to the angle at D. Wherefore, in equal circles, &c. Q. E. D.

PROP. XXVIII. THEOR. In equal circles, equal straight lines cut off equal arches, the greater equal to the greater, and the less to the less. Let ABC, DEF be equal circles, and BC, EF equal straight lines in

them, which cut off the two greater arches BAC, EDF, and the two less BGC, EHF: the greater BAC is equal to the greater EDF, and the less BGC to the less EHF.

Take (1.3.) K, L, the centres of the circles, and join BK, KC, EL, LF: and because the circles are equal, the straight lines from their

centres are equal; therefore BK, KC are equal to EL, LF; but the base BC is also equal to the base EF; therefore the angle BKC is equal (8.1.) to the angle ELF: and equal angles stand upon equal (26. 3.) arches, when they are at the centres; therefore the arch BGC is equal to the arch EHF. But the whole circle ABC is equal to the whole EDF; the remaining part, therefore, of the circumference, viz.. BAC, is equal to the remaining part EDF. Therefore, in equal circles, &c. Q. E. D.

PROP. XXIX. THEOR.

In equal circles equal arches are subtended by equal straight lines.

Let ABC, DEF be equal circles, and let the arches BGC, EHF also be equal; and join BC, EF: the straight line BC is equal to the straight line ÉF.

Take (1.3.) K, L the centres of the circles, and join BK, KC, EL, LF; and because the arch BGC is equal to the arch EHF, the angle BKC is equal (27. 3.) to the angle ELF: also because the circles ABC, DEF are equal, their radii are equal: therefore BK, KC are

equal to EL, LF; and they contain equal angles: therefore the base BC is equal (4. 1.) to the base EF. Therefore, in equal circles, &c. Q. E. D.

+ PROP. XXX. PROB.

To bisect a given arch, that is, to divide it into two equal parts.

Let ADB be the given arch; it is required to bisect it.

Join AB, and bisect (10. 1.) it in C; from the point C draw CD at right angles to AB, and join AD, DB: the arch ADB is bisected in the point D.

D

Because AC is equal to CB, and CD eommon to the trianglesACD, BCD, the two sides AC, CD are equal to the two BC, CD; and the angle ACD is equal to the angle BCD, because each of ⚫ them is a right angle; therefore the base AD is equal (4. 1.) to the base BD. equal straight lines cut off equal (28. 3.) arches, the greater equal to the greater,

But

and the less to the less; and AD, DB are each of them less than a semicircle, because DC passes through the centre (Cor. 1. 3.): Wherefore the arch AD is equal to the arch DB: and therefore the given arch ADB is bisected in D: Which was to be done.

+ PROP. XXXI. THEOR.

In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

Let ABCD be a circle, of which the diameter is BC, and centre E; draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC; the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicir cle, is less than a right angle; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle.

F

Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal (5. 1.) to EBA; also, because AE is equal to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB. But FAC, the exterior angle of the triangle ABC, is also equal (32. 1.) to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a right (7. def. 1.) angle: wherefore the angle BAC in a semicircle is a right angle.

B

E

And because the two angles ABC, BAC of the triangle ABC are together less (17. 1.) than two right angles, and BAC is a right angle, ABC must be less than a right angle; and therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle. Also because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal (22. 3.) to two right angles; therefore the angles ABC, ADC are equal to two right angles; and ABC is less than a right angle; wherefore the other ADC is greater than a right angle. Therefore, in a circle, &c. Q. E. D.

COR. From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal to the same two; and when the adjacent angles are equal, they are right angles.

PROP. XXXII. THEOR.

If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line which touches the cirele, shall be equal to the angles in the alternate segments of the circle.

Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn cutting the circle: The angles which BI) makes with the touching line EF, shall be equal to the angles in the alternate segments of the circle: that is, the angle FBD is equal to the angle which is in the segment DAB, and the angle DBE to the angle in the segment BCD.

D

From the point B draw (11. 1.) BA at right angles to EF, and take any point C in the arch BD, and join AD, DC, CB; and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line, from the point of contact B, the centre of the circle is (19.3.) in BA; therefore the angle ADB, in a semicircle, is a right (31. 3.) angle, and consequently the other two angles BAD, ABD are equal (32. 1.) to a right angle: but ABF is likewise a right angle; therefore the angle ABF is equal to the angles BAD, ABD: take from these equals the common angle ABD; and there will remain the angle DBF equal to the angle BAD, which is in the alternate ment of the circle. And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are equal (22. 3.) to two right angles; therefore the angles DBF, DBE, being likewise equal (13.1.) to two right angles, are equal to the angles BAD, BCD; and DBF has

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B