ciding with one another; then, one of the segments must fall within the other: let ACB fall within ADB, and draw the straight line BCD, and join CA, DA: and because the segment ACB is similar to the segment ADB, and that similar segments of circles contain equal angles; the angle ACB is equal to the angle ADB, the exte A rior to the interior, which is impossible. Therefore, there cannot be two similar segments of a circle upon the same side of the same line, which do not coincide. Q. E. D. PROP. XVI. THEOR. Similar segments of circles upon equal straight lines, are equal to oneanother. Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD: the segment AEB is equal to the segment CFD. For, if the segment AEB be applied to the segment CFD, so as the point A be on C, and the straight line AB upon CD, the point B shall coincide with the point D, because AB is equal to CD; therefore the straight line AB coinciding with CD, the segment AEB must (15. 3.) coincide with the segment CFD, and therefore is equal to it. Wherefore, similar segments, &c. Q. E. D. In equal circles, equal angles stand upon equal circumferences, 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 circumference BKC is equal to the circumference 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 G is equal 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 (Cor. 13. 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 (16. 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 circumference BKC to the circumference ELF. Wherefore, in equal circles, &c. Q. E. D. PROP. XVIII. THEOR. In equal circles, the angles which stand upon equal circumferences are equal to one another, whether they be at the centres or circumfe rences. Let the angles BGC, EHF at the centres, and BAC, EDF at the circumference of the equal circles ABC, DEF stand upon the equal circumferences 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 (12. 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 (14. 1.) the angle BGK equal to the angle EHF; but G H B C E F equal angles stand upon equal circumferences (17.3.) when they are at the centre; therefore the circumference BK is equal to the circumference EF: but EF is equal to BC; 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 angle BGC, and the angle at D half of the angle EHF therefore the angle at A is equal to the angle at D. Wherefore, in equal circles, &c. Q. E. D. PROP. XIX. THEOR. In equal circles, equal straight lines cut off equal circumferences, 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 circumferences 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 (3. 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; and the base BC is equal to the base EF; therefore the angle BKC is equal (7. 1.) to the angle ELF; but equal angles stand upon equal (17. 3.) circumferences, when they are at the centres: therefore the circumference BGC is equal to the circumference 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. XX. THEOR. In equal circles, equal circumferences are subtended by equal straight lines. Let ABC, DEF be equal circles, and let the circumference BGC, EHF also be equal; and join BC, EF: the straight line BC is equal to the straight line EF. Take (3. 3.) K, L, the centres of the circles, and join BK, KC, EL, LF; and because the circumference BGC is equal to the circumference EHF, the angle BKC is equal (18. 3.) to the angle ELF: and because the circles ABC, DEF are equal, the straight lines from their centres 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. XXI. PROB. To bisect a given circumference, that is, to divide it into two equal parts. Let ADB be the given circumference; it is required to bisect it. Join AB, and bisect (8. 1.) it in C; from the point C draw CD at right angles to AB, and join AD, DB: the circumference ADB is bisected in the point D. Because AC is equal to CB, and CD common to the triangles ACD, BCD, the two sides AC, CD are equal to the two BC, CD; and the angle ACD is equal to the angle BCD, because D C B to the greater, and the less to the less, and AD, DB are each of them less than a semicircle; because DC passes through the centre (Cor. 3. 3.) wherefore the circumference AD is equal to the circumference DB: therefore the given circumference is bisected in D. Which was to be done. PROP. XXII. THEOR. In a circle, the angles 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; and 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 semicircle, 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. 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 equal (5. 2.) to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a right (Def. 27. 1.) angle: where B fore the angle BAC in a semicircle is a right angle. F A D C E And because the two angles ABC, BAC, of the triangle ABC are together less (5. 2.) than two right angles, and that 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. And because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal (14. 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. PROP. XXIII. 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 touching the circle, shall be equal to the angles which are 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 BD 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. From the point B draw (9. 1.) BA at right angles to EF, and take any point C in the circumference 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 (11. 3.) in BA; therefore the angle ADB in a semicircle is a right angle, and con E BAL F |