PROP. XXVI. PROB.

In equal circles (ABC, DEF), equal angles, stand on equal circumferences or arches; whether they be at the centres (as G, H), or at the circumferences (as A, D).

G

Η

F

K

If the angles G, H be at the centre, let there be constituted the angles A, D at the circumferences on the same arches, and join BC, EF.

Because the circles ABC, DEF are equal, their semi-diameters are equal (Cor. 5. 24. 3); therefore, in the triangles BGC, EHF, the sides BG, GC are severally equal to EH, HF, and the angles G, H are equal Hyp.), therefore the bases BC, EF are equal (4. 1); but the angles BAC, EDF are equal (20. 3 and Ax. 7. 1), therefore the segments BAC, EDF are similar (Def. 11.3); and they are constituted on equal right lines BC, EF, therefore the circumferences BAC, EDF are equal (Cor. 1. 24. 3); but the whole circumference BACK is equal to the whole circumference EDFL (Hyp. & Cor. 5 & 2.524. 3), therefore the remaining circumference BKC is equal to the remaining circumference ELF.

If the equal angles, as A, D, be at the circumference, and acute; drawing BG, GC, EH and HF, the arches BKC and ELF may, in like manner, be proved equal.

But if the equal angles at the circumference be either right or obtuse, let them be bisected, their halves are equal (Ax. 7. 1), and it may be shewn as above, that the arches, on which they stand, are equal, and therefore the whole arches are equal.

PROP. XXVII. THEOR.

In equal circles (BAC, EDF), the angles which stand on equal arches (BKC, ELF) are equal, whether they be at the centres (as BGC, EHF), or at the circumferences (as BAC, EDF).

Because then, in the equal circles BAC, EDF, the angles BGK, EHF are equal [Constr.], the arches BK, EF are equal [26.3]; but the arches BC, EF are equal [Hyp.]; therefore the arches BK, BC, being each equal to EF, are equal, part and whole, which is absurd; therefore the angles BGC, EHF are not unequal, they are therefore equal; whence the angles BAC, EDF, being halves of the equal angles BGC, EHF [20. 3], are equal.

PROP. XXVIII. THEOR.

In equal circles (ABC, DEF), equal right lines (BC, EF), cut off equal arches, the greater (BAC), equal to the greater (EDF), and the less (BGC), to the less (EHF).

If the equal right lines be diameters, the proposition is manifest.

But if not, find K and L the centres of the circles [1. 3], and join BK, KC, EL, LF.

B

K

G

L

C E

F

H

Because the circles are equal, the right lines BK, KC are severally equal to EL, LF (Cor. 5. 24. 3), and BC is equal to EF (Hyp.), therefore the angles K, L are equal [8. 1], and therefore the circumference BGC is equal to the circumference EHF [26. 3]; and the whole circumference BACG is equal to the whole circumference EDFH, therefore

the remaining circumference BAC is equal to the remaining circumference EDF.

PROP. XXIX. THEOR.

In equal circles (ABC, DEF, see fig. in the preceding Prop.), equat circumferences BGC, EHF), are subtended by equal right lines (BC, EF).

If the equal circumferences be semicircles, the proposition is manifest.

If not, find the centres of the circles K, L. 3]; and, because the circumferences BGC, EHF are equal, the angles K, L are equal [27. 3]; and, because the circles ABC, DEF are equal, the right lines BK, KC are severally equal to EL, LF (Cor. 5. 24. 3); whence, the triangles BKC, ELF, having BK, KC and the included angle K, severally equal to EL, LF and the included angle L, the bases BC, EF are equal.

Scholium. What are demonstrated in the four preceding propositions about equal circles, are also manifestly true about the

same.

Cor. 1.-In equal circles [ABC, DEF, see fig. to prop. xxvi of this], sectors [BGC, EHF], which stand on equal arches [BKC, ELF], are equal.

Because the arches BKC, ELF are equal [Hyp.], the right lines BC, EF are equal [by this prop.]; and the angle BAC is equal to EDF [27. 3], and therefore the segments BAC, EDF are similar [Def. 11. 3], and, being on equal right lines BC, EF, are equal [24. 3]; taking each from the whole circles, which are equal [Hyp.j, the remaining segments BKC, ELF are equal; and the triangles BGC, EHF, being mutually equilateral, are equal Cor. 8. 1]; therefore, adding these triangles to the equal segments BKC, ELF the sectors BGC, EHF are equal.

Cor. 2.-In equal circles, equal angles, whether at the centre or circumference, are subtended by equal right lines.

For these equal angles stand on equal arches (26. 3), and the equal arches are subtended by equal right lines (by this prop.).

PROP. XXX. PROB.

To bisect a given circumference or arch of a circle (ADB).

Join AB, which bisect in C (10. 1), from C, draw CD at right angles to AB, which bisects the circumference in D.

A

D

Join AD, DB; and in the triangles ACD, BCD, the sides AC, CB are equal (Constr.), CD common to the two triangles, and the angle ACD equal to BCD (Def. 20. 1); therefore the base AD is equal to the base BD (4. 1), and therefore the circumferences AD, DB, which they subtend, are equal [28. 3], and so the given circumference ADB is bisected in D.

PROP. XXXI. THEOR.

The angle (BAC) in a semicircle, is a right angle; but an angle (as ABC), in a segment greater than a semicircle, is less than a right angle; and an angle (as ADC), in a segment less than a semicircle, is greater than a right angle.

Let E be the centre of the circle, join AE, and produce BA, as to F; and because, in the triangle EAB, the sides EA, EB are equal, the angles EAB, EBA are equal (5. 1); and because, in the triangle B EAC, EA is equal to EC, the angles EAC, ECA are equal (5. 1); therefore the whole angle BAC is equal to the two angles ABC, ACB [Ax. 2. 1]; but the

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A

D

E

exterior angle FAC, of the triangle BAC, is equal to the two angles ABC, ACB [32. 1]; therefore the angle BAC is equal to the angle FAC, and therefore a right angle [Def. 20. 1].

And, because the two angles BAC, ABC, of the triangle ABC, are less than two right angles [17. 1], and BAC is a right angle, the angle ABC, in a segment ABC greater than a semicircle, is less than a right angle.

And, because the two opposite angles ABC, ADC, of the quadrangle ABCD in the circle, are equal to two right angles [22. 3], and ABC is less than a right angle; the angle ADC, in a segment ADC less than a semicircle, is greater than a right angle.

Cor.-A circle, described about the hypothenuse [BC], of a right angled triangle [ABC], passes through the right angle [BAC]; for, if it cut the right line [BF] in any other point but [A], the angle [BAC] would be greater or less, than the angle formed at the intersection, by right lines drawn from it to the extremes of the hypothenuse BC [16. 1], which angle so formed being, by this prop. a right one; the angle [BAC] would be greater or less than a right angle, contrary to the hypothesis.

PROP. XXXII. THEOR.

If a right line (EF) touch a circle (ABCD), and from the contact (B), a right line (BD) be drawn cutting the circle ; the angles made by the tangent and cutting line, are equal to the angles in the alternate segments.

If the cutting line pass through the centre, the angles are equal, being right angles [18 and 31. 3].

If not, from the contact B, draw BA at right angles to EF [11. 1], and, having taken any point C, in the circumference BCD, join AD, DC, CB; and because BA is perpendicular to the tangent EF, the centre of the circle is in BA

[19. 3], and the angle BDA in a semicircle is a right angle [31. 3], and, therefore, in the triangle ADB, the other two angles BAD, ABD, are equal to a right angle [32. 1], and therefore to the right angle ABF; taking from each the common angle ABD, the angle DBF is equal to the angle BAD in the alternate segment. And, because, in the quadrilateral figure ABCD inscribed in a circle, the opposite angles BAD, BCD are equal to two right angles [22. 3], and therefore to the angles DBF, DBE, which are also equal to two right angles [13. 1]; taking away the equal angles BAD, DBF, the remaining angle DBE is equal to the remaining angle DCB in the alternate segment.