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EB, EC are therefore equal to one another; wherefore a E is Book III. 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 9. 3. through the other points; and the circle of which ABC is a legment is described : And 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 lefs 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 is a segment. Which was to be done.
N equal circles, equal angles stand upon equal circum
ferences, 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 : There. fore 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 to the base EF: And because the angle at A 14.1. is equal to the angle at D, the segment BAC is fimilar to the bir. def.3. segment EDF; and they are upon equal ftraight lines BC, EF; but Gimilar segments of circles upon equal straight lines are e. qual to one another ; therefore the segment BAC is equal - 14. 3. io the fegment EDF: But the whole circle ABC iş equal to the
Book III. 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.
N equal circles, the angles which stand upon equal 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 circumferences BC, EF: The angle BGC is e. qual 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 that the angle BAC is also equal to EDF : But, if not, one
a 20. 3.
of them is the greater : Let BGC be the greater, and at the point G, in the straight line BG, make b the angle BGK equal to the angle EHF; but equal angles stand upon equal circumferences o, when they are at the centre; therefore the circum ference 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 cqual circles, &e. Q. E D.
b 23. 1.
c 26. 3.
PRO P. XXVIII.
IN equal circles, equal straight lines cut off equal cir
cumferences, the greater equal to the greater, and the less to the less.
Let ABC, DEF be equal circles, and BC, EF equal straight s 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 * K, L the centres of the circles, and join BK, KC, a 1. 3. 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 to the angle ELF: But equal angles b 8. I. stand upon equal circumferences, when they are at the cen- c 26. 3. tres; 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. There fore, in equal circles, &c. Q. E. D.
IN equal circles equal circumferences are subtended by
equal straight lines.
Let ABC, DEF be equal circles, and let the circumferences
Book Ill. Take K, L the centres of the circles, and join BK, KC, MEL, LF: And because the circumference BGC is equal to the
a 1. 3.
circumference EHF, the angle BKC is equal 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 to the base EF. Therefore, in equal circles, &c. Q. E. D.
C 4. I.
O bisect a given circumference, that is, to divide it
into two equal parts.
Let ADB be the given circumference ; it is required to bi
sect it. • 10. I. Join AB, and bisect a it in C; from the point C draw CD
at right angles to AB, and join AD, DB: The circumference ADB is bifected in the point D.
Because AC is equal to CB, and CD common to the triangles ACD, BCD, the two lides AC,
A 14. I.
B c 28. 3,
straight lines cut off equal circumferences, the greater equal 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 a Cor. 1. 3. centre d; Wherefore the circumference AD is equal to the cir.
cumference DB :: Therefore the given circumference is bifected in D, Which was to be done.
PRO P. XXXI.
THE O R.
a circle, the angle in a semicircle is a right angle ;
but the angle in a fegment 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 fegment 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 to EBA ; also, because AE a s. I. 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, AZ the exterior angle of the triangle
D ABC, is equal to the two angles
b 32. I. ABC, ACB; therefore the angle BAC is equal to the angle FAC, B
C and each of them is therefore a
C IO. def. 1. right angle: Wherefore the angle BAC in a femicircle is a right angle.
And because the two angles ABC, BAC of the triangle ABC are together less d than two right angles, and that BXC d 17. 1. 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 lets than a right angle,
And because ABCD is a quadrilateral figure in a circle, any iwo of its opposite angles are equal to two right angles; tbere. e 22. 3. fore 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.
Befides, it is manifest, that the circumference of the greater segment ABC falls without the right Angle CAB, but the circumference of the less segment ÄDC falls within the right angle CAF. • And this is all that is meant, when in the