PROPOSITION XXVIII. THEOREM. In equal circles equal right lines cut off equal circumferences, the greater equal to the greater, and the less to. the less. B K G D H a 1. 3 Let ABC, DEF, be equal circles, and BC, EF, equal right lines in them which cut off the greater circumferences BAC, EDF; also the less circumferences, BGC, EHF. The greater circumference, BAC, is equal to the greater, EDF; and the less circumference, BGC, to the less, EHF. For take K, L, the centres of the circles, and join BK, KC, EL, LF. Because they are equal circles, the right lines drawn from their centres shall be equal: therefore the two BK, KC, are equal to the two EL, LF; and the base BC is equal to the base EF: wherefore the angle BKC is equal to the angle ELF. But equal angles stand b 8. 1. upon equal circumferences: wherefore the circumference BGC is equal to the circumference EHF. But the whole 26. 3. circle ABC is equal to the whole circle DEF: therefore the remaining circumference, BAC, will be equal to the remaining circumference, EDF. Wherefore in equal circles, &c. Q. E. D. Deduction. In equal circles, the greater of two chords cuts off the greater circumference. PROPOSITION XXIX, THEOREM. In equal circles the right lines are equal which subtend equal circumferences. Let ABC, DEF, be equal circles, and in them take the equal circumferences, BGC, EHF, and join BC, EF. The right line BC is equal to the right line Er. For D A K L B G F H C a find the centres K, L, of the circles, and join BK, KC, a 1. 3. EL, LF. Because, therefore, the circumference BGC is equal to the circumference EHF, the angle BKC will b 4.1. also be equal to the angle ELF, and because the circles ABC, DEF, are equal, the right lines drawn from their centres will also be equal; therefore the two BK, KC, are equal to the two EL, LF, and they contain equal angles; wherefore the base Bc is equal to the base EF." Therefore in equal circles, &c. Q. E. D. Deduction, In equal circles, the greater of two circumferences is subtended by the greater chord. PROPOSITION XXX. PROBLEM. To bisect a given circumference. Let ADB be a given circumference. It is required to bisect it. Join AB, and bisect it in c. Also from the point c draw CD at right angles to AB, and join AD, DB. Therefore because AC is equal to CB, also CD is common; the two AC, CD, are equal to the two BC, CD, and the angle ACD is equal B Wherefore the circumference AD will be equal to the circumference BD. Therefore the given circumference has been bisected. Q. E. F. PROPOSITION XXXI. THEOREM. In a circle the angle in a semicircle is a right angle, also that in a segment greater than a semicircle is less than a right angle, and that in a segment less than a semicircle is greater than a right angle, and moreover the angle of a greater segment is greater than a right angle, but that of a less segment is less than a right angle. Let ABCD be a circle whose diameter is BC, and E the centre; join BA, AC, AD, DC. The angle which is in the semicircle BAC is a right angle, also that which is in the segment ABC greater than a semicircle; viz. the angle ABC is less than a right angle, and that which is in the segment ADC, which is less than a semicircle, viz. the angle ADC, is greater than a right angle. Join AE, and produce BA to F. B F C E b с a 5. 1. Therefore because BE is equal to EA, the angle EAB will also be equal to the angle EBA. Again, because AE is equal to EC, the angle ACE will be equal to the angle CAE; therefore the whole angle BAC is equal to the two angles ABC, ACB. But the angle FAC is without the triangle ABC, and is equal to the two ABC, ACB; therefore the angle BAC is equal to the angle ↳ 16. 1. FAC; and consequently each of them is a right angle. Wherefore in the semicircle BAC, the angle BAC is a right angle. And because the two angles ABC, BAC, of the triangle ABC, are less than two right angles, but 17. 1. BAC is a right angle; the angle ABC will be less than a right angle, and it is the angle in the segment ABC which is greater than a semicircle. But since ABCD is a quadrilateral figure inscribed in a circle, also the opposite angles of quadrilateral figures are equal to two right angles, the angles ABC, ADC, will be equal to two right angles, and the angle ABC is less than a right angle; therefore the remainder ADC will be greater than a right angle, and it is in the segment ADC, which is less than a semicircle. Moreover the angle of the greater segment which is contained by the circumference ABC, and the right line AC, is greater than a right angle, but the angle of the less segment contained by the circumference ADC and the right line AC, is less than a right angle. Whence it is evident, because the angle which is contained by the right lines BA, AC, is a right angle, that which is contained by the circumference ABC and the right line AC will also be greater than a right angle. Again, because the angle contained by the right lines CA, AF, is a right angle, that which is contained by the right line ca and the circumference ADC, is less than a right angle. Therefore in a circle, &c. Q. E. D. Deductions. 1. In a right angled triangle, given the hypothenuse and perpendicular let fall from the right angle to the hypothenuse to construct the triangle. 2. If the chords of two arcs of the same circle cut each other at right angles, the squares of the four segments of the chords are, together, equal to the square of the diameter. a 19. 3. b 22.3. 3. If the diameter of a circle be divided into any two parts, and from the point of section a perpendicular be drawn to the circumference, the squares of the two parts, with twice the square of the perpendicular, shall be together equal to the square of the diameter. PROPOSITION XXXII. THEOREM. If a right line touches a circle, and from the point of contact another right line be drawn cutting the circle, the angles which this line makes with the touching line will be equal to those which are in the alternate segments of the circle. For let any right line EF touch the circle ABCD in B, and from the point в draw the right line BD anyhow, cutting the circle ABCD. The angles which BD makes with the touching line EF are equal to those which are in the alternate segments of the circle; that is, the angle FBD is equal to the angle which is in the segment DAB; also the angle DBE is equal to the angle in the segment DCB. For from the point B draw BA at right angles to EF, and in the circumference BD take any point c, and join AD, DC, CB. Therefore because any right line EF touches the circle ABCD, in the point B, and from the point of contact в a right line BA is drawn at right angles to the touching line, the centre of the circle ABCD will be in BA.a Wherefore BA is a diameter of the same circle, and ADB an angle in a semicircle is a right angle. Therefore the remaining angles BAD, ABD, are equal to a right angle. But ABF is also a right angle; wherefore the angle ABF is equal to the angles BAD, ABD. Take away the common angle ABD. Therefore the remainder DBF is equal to that which is in the alternate segment of the circle; namely, the angle BAD. And because ABCD is a quadrilateral figure inscribed in a circle, and its opposite angles are equal to two right angles, the angles DBF, DBE, are equal to the angles BAD, BCD, of which DBF is shown to be equal to BAD. Wherefore the remainder DBE will be equal to DCB, viz. to that which is in the alternate segment of the circle DCB. If, therefore, any right line, &c. Q. E. D. b E B Deductions. 1. If a right line be drawn a tangent to an arc at the point of bisection, it shall be parallel to the chord of the arc. 2. If a triangle be described in a circle, and from the vertex a line be drawn touching the circle, the angles formed by this line, and the two sides of the triangle, shall be respectively equal to the three angles of the triangle. PROPOSITION XXXIII. PROBLEM. Upon a given right line to describe a segment of a circle which shall contain an angle equal to a given rectilineal angle. Let AB be a given right line, and c a given rectilineal angle. It is required upon the given right line AB to describe a segment of a circle which shall contain an angle equal to the angle at c. At the right line AB, and at the given point A in it, make the angle BAD equal to the angle c, and from the point a draw AE at right angles to AD. But bisect AB in F, and from the point F draw FG at B F a 23. 1. A right angles to AB, and join GB. Therefore because |