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the Angle at the Circumference, when the same
ET ABC be a Circle, at the Center whereof
For join AE and produce it to F.
Then because EA is equal to EB, the Angle EAB shall be equal to the Angle EBA*. Therefore the * 5. 1. Angles EAB, EBA, are double to the Angle E AB; but the Angle BeF is t equal to the Angles EAB, † 32. 1. EBA; therefore the Angle BEF is double to the Angle E AB. For the same Reason, the Angle FEC is double to E AC. Therefore the whole Angle BEC is double to the whole Angle BAC. Again, let there be another Angle BDC, and join DE, which produce to G. We demonstrate in the same Manner, that the Angle GEC is double to the Angle GDC; whereof the Part GEB is double to the Part GDB. And therefore BEC is double to BDC. Consequently, an Angle at the Center of a Circle is double to the Angle at the Circumference, when the fame Arc is the Base of the Angles; which was to be demonstrated.
are equal to each other.
ET ABCDE be a Circle, and let BAD, BED,
be Angles in the same Segment BAED. I say, those Angles are equal.
For let F be the Center of the Circle ABCDE, and join BF, FD.
Now because the Angle BFD is at the Center, and the Angle BAD at the Circumference, and they
stand upon the fame Arc BCD; the Angle BFD * 20 of this. will be * double to the Angle BAD. For the same
Reason, the Angle BFD is also double to the Angle
If the Angles BAD, BED, are in a Segment less
than a Semicircle, let AE be drawn; and then all + 32. I.
the Angles of the Triangle ABG aret equal to all
proved, and the Angles AGB, DGE, are also eI 15. 1. qual I, for they are vertical Angles. Wherefore the
remaining Angle BAG is equal to the remaining An-
ET ABDC be a Circle, wherein is described the
quadrilateral Figure ABCD. I say, two oppo-
For join AD, BC.
are equal to two Right Angles. But the Angle ABC + 21 of tbis. is t equal to the Angle ADC; for they are both in
the same Segment ABD C. And the Angle ACB is
* equal to the two Right Angles. Therefore likewise, * 32, 1. the Angles BAC, BDC Thall be equal to two Right Angles. And after the same way we prove, that the Angles ABD, ACD, are also equal to two Right Angles
. Therefore the opposite Angles of any Quadrilateral Figure described in a Circle, are equal to two Right Angles ; which was to be demonstrated.
cannot be set upon the same Right Line, and
on the same Side thereof. FORCE OR if this be possible, let the two similar and un
equal Segments ACB, ADB, of two Circles stand upon the Right Line AB on the fame Side thereof. Draw ACD, and let CB, BD, be joined. Now because the Segment ACB is similar to the Segment AD B, and similar Segments of Circles are * such * Def
. 11. which receive equal Angles; the Angle ACB will of this. be equal to the Angle AĎ B, the outward one to the inward one; which is † absurd. Therefore fimilar † 16. Is and unequal Segments of two Circles, cannot be set upon the same Right Line, and on the fame Side thereof; which was to be demonstrated,
Lines, are equal to one another.
standing upon the equal Right Lines AB, CD. I fay, the Segment AEB is equal to the Segment CFD.
For the Segment A E B being applied to the Segment CFD, so that the Point A co-incides with C, and the Line AB with CD; then the Point B will co-incide with the Point D, fince AB and CD are equal. And since the Right Line AB co-incides with
CD, the Segment AEB will co-incide with the Segment CFD. For if at the same Time that AB co-incides with CD, the Segment AEB should not co-incide with the Segment CFD, but be otherwise, as CGD; then a Circle would cut a Circle in more
Points than two, viz. in the Points C, G, D; which * 10 of thi. is * impoffible Wherefore if the Right Line AB
co-incides with CD, the Segment AEB will co-incide with and be equal to the Segment CFD. Therefore similar Segments of Circles being upon equal Right Lines, are equal to one another; which was to be demonstrated.
Circle whereof it is the Segment.
1 23. 1.
* 6. 1.
ET AB C be a Segment of a Circle given. It L is required to describe a Circle, whereof ABC is a Segment.
Bisect * AC in D, and let DB be drawn from the Point D at Right Angles to AC, and join AB. Now the Angle ABD is either greater, equal, or less than the Angle BAD. And first let it be greater, and make I the Angle BAE at the given Point A, with the Right Line BA, equal to the Angle ABD; produce DB to E, and join E C.
Then because the Angle ABE is equal to the Angle BAE, the Right Line B E will be * equal to EA. And because AD is equal to DC, and D E common, the two Sides AD, DE, are each equal to the two Sides CD, DE; and the Angle ADE is equal to the Angle CDE; for each is a Right one. Therefore the Dafe A E is equal to the Base EC. But A E has been proved to be equal to EB. Wherefore BE is also equal to E C. And accordingly the three Right Lines AE, EB, EC, are equal to each other. Therefore a Circle described about the Center E, with either of the Distances A E, EB, EC, shall pass thro' the other Points, and be that required to be described. But it is manifest that the Segment ABC is less than
a Semicircle, because the Center thereof is without the same.
But if the Angle ABD be equal to the Angle BAD; then if AD be made equal to BD, or DC, the three Right Lines AD, DB, DC, are equal between themselves, and D will be the Center of the Circle to be described ; and the Segment ABC is á Semicircle.
But if the Angle ABD is less than the Angle BAD, let the Angle BAE be made, at the given Point A with the Right Line BA, within the Segment ABC, equal to the Angle ABD.
Then the Point E, in the Right Line D B, will be the Center, and ABC a Segment greater than a Semicircle. Therefore a Circle is described, whereof a Segment is given; which was to be done.
THE ORE M.
Circumferences, whether they be at their Centers,
For let BC, EF, be joined. Because ABC, DEF, are equal Circles, the Lines drawn from their Centers will be equal. Therefore the two Sides BG, GC, are equal to the two Sides EH, HF; and the Angle G is equal to the Angle at H. Wherefore the Base BC is * equal to the Base EF. Again, because the * 4. I. Angle at A is equal to that at D, the Segment BAC will be + fimilar to the Segment EDF; and they are t Def, 11. upon equal Right Lines B C, EF. But those similar Segments of Circles, that are upon equal Right Lines, are equal to each other. Therefore the Segment # 34 of tbis. BAC, will be † equal to the Segment EDF. But † Defi 1 the whole Circle ABC, is equal to the whole Circle PEF. Therefore the remaining Çircumference BKC,