Right Angle. Therefore the Angle FDB is equal to Right Line into two equal Parts, and at Right An- ting Line. PROPOSITION II. THEOREM. of a Circle, the Right Line joining those two a which let any two Points A, B, be assumed. I fay, a Right Line drawn from the Point A to the Point B, falls within the Circle. For let any Point E be taken in the Right Line AB, and let DA, DE, DB, be joined. Then because DA is equal to DB, the Angle DAB will be * equal to the Angle DBA; and since the * 5. I. Side A E of the Triangle DAE is produced, the Angle DEB will be + greater than the Angle DAE, † 16. 1. but the Angle DAE is equal to the Angle D BE; therefore the Angle DEB is greater than the Angle DBE. But the greater Side fubtends the greater Angle. Wherefore DB is greater than DE. But D B only comes to the Circumference of the Circle; therefore DE does not reach so far. ' And so the Point E falls within the Circle. Therefore, if two Points are af fumed in the Circumference of a Circle, the Right Line joining those two. Points' fall fall within the Circle, which was to be demonstrated. Coroll. Hence if a Right Line touches a Circle, it will touch it in one Point only, PRO PROPOSITION III. THEOREM. ter, cuts any other Right Line not drawn throm CD, drawn thro' the Center, bisects the Right it at Right Angles. • of this. For * find E the Center of the Circle, and let EA, EB, be joined. Then because AF is equal to FB, and F E is common, the two Sides AF, FE are equal to the two Sides BF, FE, each to each, but the Base EA is equal to the Base EB. Wherefore the Angle AFE + 8. s. fhall be equal to the Angle BFE. But when a Right Line standing upon a Right Line makes the adjacent Angles equal to one another, each of the equal Angles | Def. 10. 1. is I a Right Angle. Wherefore AFE, or BFE, is a Right Angle. And therefore the Right Line CD drawn thro' the Center, bisecting the Right Line AB not drawn thro' the Center, cuts it at Right Angles. Now if CD cuts AB at Right Angles, I say, it will bisect it, that is AF will be equal to FB. For the same Construction remaining, because EA, being drawn from the Center, is equal to E B, the Angle EAF shall be * equal to the Angle EBF. But the Right Angle AFE is equal to the Right Angle BFE; therefore the two Triangles EAF, EBF, have two Angles of the one equal to two Angles of the other, and the Side EF is common to both. Wherefore the other Sides 4 26. . of the one shall be f equal to the other Sides of the other: And so AF will be equal to FB. Therefore if in a Circle a Right Line drawn thró the Center, cuts any other Right Line not drawn thrò the Center, into two equal Parts, it shall cut it at Right Angles; and if it cuts it at Right Angles, it shall cut it into two equal Parts ; which was to be demonstrated. PRO PROPOSITION IV. THEOREM. tbro' the Center, cut each other, they will not cut each other into two equal Parts. L ET ABCD be a Circle, wherein two Right Lines AC, BD, not drawn thro' the Center, cut each other in the Point E. I say, they do not bisect each other. For, if possible, let them bisect each other, so that AE be equal to EC, and BE to ED. Let the Cen * I of this. ter F of the Circle ABCD be * found, and join EF. Then because the Right Line F E drawn thro' the Center, bisects the Right Line AC not drawn thro' the Center, it will + cut AC at Right Angles. And t 3 of tbis. so FEA is a Right Angle. Again, because the Right Line FE bisects the Right Line BD not drawn thro the Center, it will + cut BD at Right Anglos. Therefore FEB is a Right Angle. But FEA has been shewn to be also a Right Angle. Wherefore the Angle FEA will be equal to the Angle FEB, a less-to a greater ; which is abfurd. Therefore AC, BD, do not mutually bifect each other. And so if in a Circle two Right Lines, not being drawn thro' the Center, cut each other, they will not cut each other into two equal Parts; which was to be demons strated. PROPOSITION V. THE O R E M. have the same Center. L ET the two Circles ABC, CDG, cut each other in the Points B, C. I say, they have not the fame Center. For if they have, let it be E, and join E C, and draw EFG at pleasure. F Now Now because E is the Center of the Circle ABC, CE will be equal to EF. Again, because E is the Center of the Circle CDG, ČE is equal to E G. But CE has been shewn to be equal to EF. Therefore E F shall be equal to EG, a less to a greater, which cannot be. Therefore the Point E is not the Center of both the Circles ABC, CDG. Wherefore, if two Circles cut one another, they shall not have the same Center; which was to be demonstrated. PROPOSITION VI. THE ORE M. If two Circles touch one another inwardly, they will not have one and the same Center. ET two Circles ABC, CDE, touch one an other inwardly in the Point C. I say, they will not have one and the same Center. For if they have, let it be F, and join F C, and draw F B any how. Then because F is the Center of the Circle ABC, CF is equal to FB. And because F is only the Center of the Circle CDE, CF shall be equal to EF. But CF has been shewn to be equal to FB. Therefore FE is equal to FB, a less to a greater ; which cannot be. Therefore the Point F is not the Center of both the Circles ABC, CDE. Wherefore, if two Circles touch one another inwardly, they will not have one and the fame Center; which was to be demonstrated. PRO PROPOSITION VII. THEOREM in which assume some Point F, which is not the For let BE, CE, GE, be joined. Again, because BE is equal to CE, and FE is Again, because GF and FE are * greater than * 20. 1. GE, and GE is equal to ED; GF and F E shall be greater than ED; and if FE, which is common, be taken away, then the Remainder GF is greater than the Remainder FD. Wherefore FA is the greate.t 20. I. |