d

58.

2

8. EM FM, and MD common to both As, EMD, FMD, the Bafe DEr Bafe DF. By the fame Method of Reasoning DF r DC. Hence we have fhewn that DA r DE, DEr DF, and DF r DC. Q. E. D.

70.

2

Second. Because MK + KD°r MD, taking away 18. equals MK, MG, there will remain KD г DG. And because the 4 MKD is within the ▲ MLD, the two Sides MK, KD, taken together must be a the two Sides ML, LD, which include them, (for it is manifeft a Thing containing must be r the Thing contained ;) and taking away equals MK,

ML, the Remainder KD, A the Remainder LD. In the fame Manner DL is A DH; . DG is the leaft, DK A DL, and DL A DH. Q. E. D.

Thirdly. Let the DMB DMK; then we affirm that the Line DB DK, and that no other. Line DN can be DK: For, that DB = DK, 18. may be thus demonftrated. Because MK2 MB, MD common, and z DMB DMK by the Sup58. pofition, the Base DB Base DK. Q. E. D.

45.

Pl.3.F.z.

134.

e

d

And that no other Line DN can be DK, may be thus fhewn: DK is = DB, as has been just proved; ..if DN be also DK, then will DN be DB, that is, the nearer to the least = the more remote, which is contrary to what has been above demonftrated, and confequently impoffible: DN is not DB. Q. E. D. =

136. THEOREM 4. If a Point D be taken within a Circle, from which there fall more than two equal Lines to the Circumference, viz. DA, DB, DC, that Point D is the Center of the Circle.

For if it is poffible that D is not the Center, let any other Point E be the Center, in the Diameter FDEG. Then, if D be not the Center, DG is the greatest Line from it to the Circumference, and DC

г DB,

1

r DB, and DB r DA; but they are alfo equal by the Suppofition, which is impoffible : ... E is not the Center of the Circle ABC, and in the fame Manner it may be demonftrated that no other Point but D is the Center. QE. D.

137. THEOREM 5.

in more than two Points.

One Circle cannot cut another

For if it be poffible, let the Circle ABC cut the pl. III. Circle DFF in more than two Points, viz. in B, G, F. 3. F; and let K be the Center of the Circle ABC, and BF, GK, FK, be joined; then, because within the Circle DEF there is taken the Point K, from which to the Circumference DEF fall more than two equal Lines, KB, KG, KF, the Point K must be the Center of the Circle DEF; but it is also the 136. Center of the Circle ABC, and fo the two Circles ABC, DEF, would have the fame common Center and Radii, and confequently cannot cut each other, but would coincide, and . properly properly speaking, would be but one Circle: ...one Circle cannot cut another in three Points. Q. E. D.

138. A LEMMA, or THEOREM 6. If two right- Pl. III. angled Triangles have two Sides of one, AC, CB, F. 4. equal to two Sides of the other, DF, FE, each to each (viz. AC = DF, CB FE;) the remaining Side of one AB, will be the remaining Side of the other DE.

AC2

For AB + BC AC, and DE + EF =
DF2: But ACDF by the Suppofition,
= DF2, ... AB2 + BC2 = DE2 + EF2; but
BC EF by the Suppofition, and fo BC

b

EF1,

taking away thefe equals, there remains AB — DE2, and confequently ABDE. Q. E. D.This Demonftration proves the Theorem to be true, when the Side oppofite to the Rt. and one of the other Sides of one, are the Side oppofite to the Rt.

15.

45.

47.

Pl. III.

F. 5.

b

and one of the other Sides of the other A. But it is equally true when the two Sides containing the Rt. in one the two Sides containing the in the other: for it then becomes the fame as

Rt.
Article 58.

*138. THEOREM 7. Equal Lines, AB, CD, in the Circle ABDC, are equally difiant from the Center E; and thofe AB, CD, equally distant from the Center, are equal to each other.

a

First, that if AB CD, they are equidiftant from the Center, may be thus demonftrated. Let EF be to AB, EGL to CD, and join AE and CE. Then the LS AFE and CGE being Rt. 4s, the As AFE, CGE, are Rt. 4 As; and AB, 134. CD, are bifected by the Lines EF, EG; and as their Wholes are equal, their Halves must be equal alfo, that is, AF CG. Hence, in the two Rt. As AFE, CGE, are two Sides of one AE, AF, respectively two Sides of the other CE, CG, (for AE, CE, are Radii of the fame Circle and .. 138, equal,) the remaining Side FE of one is the remaining Side GE of the other; which is all that is meant by faying, the Lines AB, CD, are equidiftant from the Center. Q. E. D.

c

Secondly. If AB, CD, are equidiftant from the Center, it may be easily demonstrated that they are each other. For AE being CE, and EF = EG, by the Suppofition, the Rt. As AFE, CGE, have two Sides of one AE, EF,

c

two Sides CG; but respective

138. of the other, CE, EG, and .: AF 134 AF, CG, are the Halves of their Wholes AB, CD; and as the Halves AF, CG, are proved to be equal to each other, their Wholes must be fo too; that is, AB CD. Q. E. D.

Pl.3.F.6.

139. THEOREM 8. The Diameter AD is the greateft Line in a Circle ABCD; and of any others BC, FG,

that