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PROPOSITION IV. THEOREM III. F in a circle (ADCB) two chords (AC, DB,) cut one another, they are divided into two unequal parts. Hypothesis.
Thefs. The two chords AC, DB, of the O ADCB
These chords are divided ina cut one another in the point E.
to iwo unequal parts,
ECAUSE the diameter, or its part FE, bisects each of the chords AC, DB, of the O ADCB (Sup.). 1. This straight line FE is I upon each of the chords AC, DB.
P. 3. B. 1. 2. Consequently, the V FEB, FEA, are = to one another; which s Ax.10.B.i. is impossible.
Ax. 8.B.1. 3. Wherefore, the two chords AC, DB, are divided into two unequal parts.
Which was to be demonftrated,
PROPOSITION V. THEOREM IV. F two circles (ABE, ADE,) cut one another, they shall not have the same center (C). Hypothesis.
Thesis. ABE, ADE, are two o which cut
Those two have different one another in the points A & E.
} Pos. 1.
2. And from the same point C, draw the straight line CB ; which
cuts the two o in D & B.
ECAUSE the straight lines CA, CD, are drawn from the center C to the O ADE (Prep. 1. & 2.). 1. These straight lines CA, CD, are = to one another.
D. 15. B. 1. It is proved in the same manner, that : 2. The straight lines CA, CB, are = to one another. 3. Consequently, CB will be to CD, which is impossible.
Ax.8. B. I. 4. Therefore, the two circles ABE, ADE, have not the same center.
Which was to be demonstrated.
PROPOSITION VI. THEOREM V. F two circles (BCA, ECD,) touch one another internally in (C); they shall not have the same center (F). Hypothefis.
Thesis. The OECD touches the O BCA
Those two o bave different internally in C.
ECAUSE the point F is the center of the O BCA (Sup.). 1. The rays FB, FC, are = to one another.
Again, the point F being also the center of O ECD (Sup.) 2. The rays FE, FC, are = to one another, 3. Consequently, FB = FE (Ax. 1. B. 1.); which is impossible. 4. Wherefore, the two O BCA, ECD, have not the same center.
Which was to be demonstrated.
PROPOSITION VII. THEOREM VI. F any point (F) be taken in a circle (AHG) which is not the center (E), of all the straight lines (FA, FB, FC, FH,) which can be drawn from it to the circumference, the greatest is (FA) in which the center is, & the part (FD) of that diameter is the least, & of any others, that (FB or FC) which is nearer to the line (FA) which passes thro' the center is always greater than one (FC or FH) more remote, & from the same point (F) there can be drawn only two straight lines (FH, FG,), that are equal to one another, one upon each side of the shortest line (FD). Hypothesis.
Thesis. I. The point F taken in the O AHG is 1. FA is the greatest of all the not the center E.
ftraight lines which can be drawn
from the point F to the O AHG. II. The straight line FA, drawn from II. FD is the least.
the point F, passes thro' the center E III. And of any others FB or FC which of the O AHG.
is nearer to FA is > FC or FH more
remote. ļII. And the straight lines FB, FC, FH, IV. From the point F there can be drawn
are drawn from the point F to the only two = Efraight lines FH, FG, O AHG.
one upon each side of the fortefFD.
1. Preparation. Draw the rays EB, EC, EH, &c. Fig. 1.
DEMONSTRATION. HE two sides FE + EB of the A FEB are > the third FB. P. 20. B. 1, But EB is = to EA (D. 15. B. 1.). 2. Therefore, FE + EA, or FA is > FB.
It is proved in the fame manner that: 3. The straight line FA, is the greatest of all the straight lines drawn from the point F to the O AHG.
Which was to be demonstrated. I.
4. Again, the two fides FE + FH of the A FEH are the third EH, P.ZÓ, B. 1,
And ED being = to EH (D. 15. B. 1.).
5. The straight lines FE + FH are also > ED.
Therefore, taking away from both sides the part FE: 6. The straight line FH will be > FD; or FD <FH,
Ax.5. B. . It is proved in the same manner that : 7. The straight line FD, which is the produced part of PA, is the least of all the straight lines drawn from the point F to the O AHG.
Which was to be deinonstrated. II.
Moreover, the side FE being common to the two A FEB, FEC, the side EB = the side EC (D. 15. B. 1.), & the V FEB >
V FEC (Ax. 8. B. 1.). 8. The base FB will be the base FC.
P. 24. B. 1. For the same reason : 9. The straight line FC is > FH. 16. Consequently, the straight line FB or FC which is nearer the line FA, which passes thro' the center, is > FC or FH more remote.
Which was to be demonstrated. III.
II. Preparation. Fig. 2.
P. 23. B.L 2. From the point F to the point G, draw the straight line FG. Pofi 1, Then, EF being common to the two A FEH, FEG, the side EH =the side EG (D. 15. B. 1.), & the V FEH= to the V FEG
(II. Prep. 1.). 11. The base FH will be = to the base FG.
P. 4. B.1. But because any other straight line, different from FG, is either
nearer the line FD, or more remote from it, than FG. 12. Such a straight line will be also < or > FG (Arg. 10.). 13. Wherefore, from the faine point F, there can be drawn only two
straight lines FH, FG, that are = to one another, one upon each side of the shortest line FD.
Which was to be demonstrated. IV.