a

C

Let CF be || AD or BE, and let AH be CB; and let HK be a to AB or DE. Then, because AB1 = AD, by, taking away equals CB, AH, there remains AC DH. Because HG, DF are Parallels, alfo HD || GF, the oppofite 4s of the Figure HGFD are equal to each other; but the D is a Rt. 4. the other 3 are Rt. 4s, and fo HGFD is a Square each Side AC. By the fame Reason CBKG is a Square each Side = CB; and by the like Method of Reasoning ACGH and KEFG are Rectangles whofe Sides are AC and CB. But the Squares DG, GB, and two Rectangles AG, GE, make up the whole Square ABED. Q. E, D,

a

120. COROLLARY 1. Hence the Square of any Line is equal to four Times the Square of half that Line.

121. THEOREM 2. A Rectangle made by the Sum and Difference of two right Lines is equal to the Difference of the Squares of the faid Lines.*

Let AB, AC, represent the two Lines, and AK, AG, their Squares. Let HG be produced to HF, till GF AB, or, which is the fame, till HF = AC + AB. Let FE be to AH, and let IK be produced to meet FE in E. Then, becaufe HF = AC+ AB, the Sum of the Lines, and IK = AH —AI ACAB the Difference of the Lines, 'tis plain the Rectangle HE is contained under the Sum HF, and Difference IK, of the Lines AC, AB,

34.

47.

€ 98.

a 34.

Pl. 2. Fig.

47.

a

b

=

- CD

AB, which we are now to fhew to be the Gnomon HDB, the Difference of the Squares AK, AG; which may be easily done, for GF being = AB by 34 the Suppofition, BK, and DG (= CG ACAB) BC, the Rectangles BD, DF, € 100. are contained under equal Sides, and are ·.·• ‘equal to each other; . to each adding the Rectangle HD common, we have BD + DHDF+ HD, that is, the Gnomon HD the Rectangle HE. Q. E. D.

46.

122. These two Theorems are fufficient for our prefent Purpose; we fhall therefore pafs on to the third Book. We intend hereafter to give algebraic Demonftrations to all the Theorems of this Book.

123. Def. 1.

E

BOOK III.

QUAL Circles are those whofe Diameters are equal, or from whose Centers the Lines to the Circumferences are equal : Or, in other Words, they are those that have equal Radii.

124. Scholium. This is not, frictly speaking,`a Definition, but a Theorem, whofe Truth is manifeft; for if the Circles are laid on one another fo that their Centers may coincide, the Circumferences will alfo coincide, as the Radii are equal, and ·.· the Circles will exactly cover; and be. other.

= each

125. Definition 2. A Line is faid to touch a Circle when it meets the Circle in one Point, and being produced does not touch it in any other Point: And fuch Line is called a Tangent to the Circle.

126. Definition 3. Circles are faid to touch one another, which meet, but do not cut each other.

127. Definition 4.

Lines are faid to be equally distant from the Center of a Circle, when the Perpendiculars drawn to them from the Center are equal.

128. Def. 5. But the Line on which the greater Perpendicular falls, is faid to be further from the Center.

129. Def. 6. A Segment of a Circle is the Figure contained by a Line and the Circumference it cuts off, (fee Pl. 2. F. 16.) and the Line is called: the Chard.

130. Def. 7. The Angle in a Segment is that which is contained by two Lines drawn from any Point in the Circumference (or Arch) of the Segment to the Extremities of the Chord Line, or Base of the Segment. See Pl. 2. F. 17.

131. Def. 8. And an Angle is faid to ftand upon the Circumference which the Lines that contain the Angle intercept between them.

132. Def. 9. The Sector of a Circle is the Figure contained by two Lines drawn from the Center, and the Circumference between them. See Pl. 2. Fig. 18.

133. Def. 10. Similar Segments are thofe in which the Angles are equal, or which contain equal Angles.

134. THEOREM I. If a Line CD, drawn through F1.2.F.19. the Center E of a Circle, bifect any Chord AB, it will cut it at right Angles. And if the Diameter CD is at right Angles to the Chord AB, it will bifect it.

For

a

For firft, Let EA, EB, be joined; then, becaufe 18. AF FB by the Suppofition, and AE EB being Radii, and FE common, the As AEF, BEF, have three Sides of one the three Sides of the other, 95. each to each, and they are equal in all Refpects; confequently the AFE BFE, and ca Rt. 4. Q. E. D.

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€ 14.

each =

Secondly. But if CD is at Rt. 4s to AB, then, 59. fince AE EB, the EAF4EBF, but AFE BFE, each being a Rt. 4 per Art. 14. Hence, in the As AFE, BFE, are two Ls and one Side of one two ▲s and one Side of the other, each to each, and the As are equal in all Refpects Confequently, their Bafes FA, FB, are equal to each other. Q. E. D.

94.

Pl.2.F.20.

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134. THEOREM 2. If any Point F, which is not the Center, be taken in the Diameter AD of the Circle ABCD, whofe Center is E; then, of all the Lines FB, FC, FG, &c. that can be drawn from F to the Circumference, that FA, in which the Center is, is the longest, and the other Part FD of that Diameter AD is the leaft; and of the others, that which is nearer to the Line FA is longer than one more diftant; thus, FB greater than FC, and FC greater than FG. And from the fame Point F there can be drawn to the Circumference only two Lines equal to each other, viz. one on one side and the other on the other Side of the Diameter AD, making equal Ls with it.

For ift. Let BE, CE, GE, he joined, then be* 70. cause two Sides of any A are r the third, BE + 18. EFT BF; but BEAE, AE + EF (= ·.· BEEF, and .) гBF. Again, BE being' = CE, EF, common, and ▲ BEFгZCEF, the Bafe BF of the ABEF is r the Bafe CF of the ACEF; and for the fame Reason the Bafe CF r Bafe GF. Again, because GF+ FE is r GE, and GE =

€ 92.

ED,

Hence
BFгCF, CFr

ED, GF+FE must be r ED, and taking away
the Part FE common, GF must be rFD.
we have proved that AF F BF,
GF, and GFr FD: Which were to be demonftrated.

Secondly. Let DFH be an 4

GED, then we

and that no other

FG may be thus

d 58.1

affirm the Line FH is FG, can be equal to it; for that FH fhewn. Becaufe GE EH, EF common, and 18. 4 GED DEF by the Suppofition, the Base GF Bafe FH. Q. E. D. And that no other Line can be FG, we thus prove. For if poffible, let FK be a Line FG; and FH has been just fhewn to be FG, and . FK would be = FH, that is, the two Lines FK, FH, each other, though one is more remote from the Line which paffes through the Center than the other; which is contrary to what has been proved above, and ·.· FK is not FH, and cannot be to FG.

Q. E. D.

135. THEOREM 3. Let ABC be a Circle, and D any Point without it, from which Lines are drawn to the Circumference. Then we affirm, that of those which fall upon the concave Part of the Circumference AEFC, the greatest fhall be DA, which paffes thro' the Center M; and the nearer to it greater than the more remote, viz. DE r DF, and DF F DC. But of thofe which fall on the convex Part of the Circumference, the leaft is DG, which coincides with the Diameter AG produced; and the nearer to it is always leffer than the more remote, viz. DK A DL, and DLA DH. And only two equal Lines can be drawn from the Point D to the Circumference, one upon each Side of the leaft.

b

For, ift. Let the Points ME, MF, MC, MK, ML, MH, be joined; then AM EM, . adding MD common, EM + MD (= AM + MD) AD; but EM + MD r DE,. alfo DA F DE. Again, because EMD г 4 FMD,

F

c

EM

e 45