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of the angle DAB. For the same reason the angle EDC is Book III. double of the angle DAC. Therefore the whole angle BDC is double of the whole angle BAC.
N. Again, let D, the centre of the circle, be without the angle BAC, and join AD, and produce it to E. It may be demonstrated, as in the first case, that the angle EDC is double of the angle DAC, and that EDB, a part of the first, is double of DAB, a part of the other; therefore the re
E maining angle BDC is double of the remaining angle BAC.
B Therefore, the angle at the centre, &c. Q. E. D.
PROP. XXI. THEOR.
THE angles in the same segment of a circle are equal to one another.
Let ABCD be a circle, and
E BAD, BED, angles in the same segment BAED; the angles BĂD, BED are equal to each other
Take F, the centre of the circle ABCD. Let the segment BAED be greater than a semicircle, and B join BF, FD. The angle BFD at the centre is doublea of each of the angles BAD, BED at the circumference, which have the same part of the circumference BCD for their base; therefore the angle BAD is equal to the angle BED.
Again, let the segment BAED be not greater than a semicircle, and let BAD, BED be angles in it; these also are
a 20. 3.
a 21. 3.
THE opposite angles of any quadrilateral figure described in a circle are together equal to two right angles.
Let ABCD be a quadrilateral figure in the circle ABCD; any two of its opposite angles are together equal to two right angles. Join AC, BD.
The angle CAB is equal a to the angle CDB, because they are in the same segment BADC; and the angle ACB is equal to the angle ADB, because they are in the same segment ADCB; therefore the whole angle ADC is equal
B to the angles CAB, ACB. To each of these equals add the angle ABC; and the angles ABC, ADC are equal to the an-' gles ABC, CAB, BCA. But ABC, CAB, BCA are equal to two right anglesb; therefore also the angles ABC, ADC are equal to two right angles. In the same manner the angles BAD, DCB may be shown to be equal to two right angles. Therefore, the opposite angles, &c. Q. E, D.
PROP. XXIII. THEOR.
UPON the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with each other.
If it be possible, let the two similar segments of circles,
a 10.3. other in
any other pointa ; one of the segments must therefore fall within the other. Let ACB fall within ADB; draw the straight line BCD,
B and join CA, DA. Because the segment ACB is similar to the segment ADB, the angle ACB is equalb to the angle ADB, the exterior to the interior, b 9. Def. 3. which is impossible. Therefore, upon the same straight c 16. 1. line, &c. Q. E. D.
PROP. XXIV. THEOR.
SIMILAR segments of circles, upon equal straight lines, are equal to one another.
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 that the point A may be on C, and the straight line AB upon CD, the point B will coincide with the point D, because
a 23. 3.
AB is equal to CD. Therefore the straight line AB coinciding with CD, the segment AEB musta coincide with the segment CFD, and therefore is equal to it. Wherefore, similar segments, &c. Q. E. D.
PROP. XXV. PROB.
A SEGMENT of a circle being given, to describe the circle of which it is the segment.
b 11. 1.
c 6. 1. d 9. 3.
Let ABC be the given segment of a circle ; it is required to describe the circle of which it is the segment.
Bisecta AC in D, and from the point D drawb DB at right angles to AC, and join AB. First, let the angles ABD, BAD be equal to each other; then the straight line BD is equal to DA or DC; and because the three straight lines DA, DB, DC are all equal, D is the centre of the circled. From the centre D, at the distance of any of the three lines DA, DB, DC, describe a circle; this will pass through the other points, and the circle of which ABC is a segment is described. Because the centre D is in AC, the segment ABC is a semicircle.
If the angles ABD, BAD are not equal to each other, at the point A, in the straight line AB, make the angle BAE equal to the angle ABD, and produce BD, if necessary, to E, and join EC. Because the angle ABE is equal to the angle BXE, the straight line BE is equal to EA ; and be
e 23. 1.
cause AD is equal to DC, and DE common to the triangles Book III. 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 to the base EC. But AE was shown to f 4. 1. be equal to EB, wherefore also BE is equal to EC. . Therefore the three straight lines AE, EB, EC are equal to one another; wherefored E is the centre of the circle. From the centre E, at the distance of any of the three lines AE, EB, EC, describe a circle ; this will pass through the other points, and the circle of which ABC is a segment is described. If the angle ABD be greater than the angle BAD, the centre E falls without the segment ABC, which therefore is less than a ser icircle; but if the angle ABD be less than BAD, the centre E falls within the segment ABC, which therefore is 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.
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.