Sidebilder
PDF
ePub
[ocr errors]

PROPOSITION 5.-THEOREM.
If two circles cut one another, they shall not have the same centre.

(Reference-I. def. 15.)
Hypothesis.—Let the two circles ABC, CDG, cut one
another in the points B, C.

Sequence.—The circles ABC, CDG, shall not have the same centre.

Construction.-For, if it be possible, let E be their centre; join EC, and draw any straight line EFG, meeting the circles in F and G.

Demonstration.-1. Because E is assumed to be the centre of the circle

A ABC, EC is equal to EF. (I. def. 15.)

2. And because E is assumed to be the centre of the circle CDG, EC is equal to EG.

3. But EC was shown to be equal to EF;

4. Therefore EF is equal to EG (ax. 1), the less to the greater, which is impossible;

5. Therefore E is not the centre of the circles ABC, CDG. Conclusion.—Therefore, if two circles, &c. 2. E. D.

E

PROPOSITION 6.--THEOREM.
If one circle touch another internally, they shall not have the

same centre.

(Reference-I. def. 15.) Hypothesis.—Let the circle CDE touch the circle ABC internally in the

point C. Sequence.—The circles ABC, CDE, shall not have the same centre.

Construction.-For, if it be possible, let F be their centre; join FC, and draw any straight line FEB, meeting the circles in E and B.

Demonstration.-1. Because F is assumed to be the centre of the circle ABC, FC is equal to FB. (I. def. 15.)

2. And because F is assumed to be the centre of the circle CDE, FC is equal to FE.

3. But FC was shown to be equal to FB;

[ocr errors]

A

B

F

E

4. Therefore FE is equal to FB (ax. 1), the less to the greater, which is impossible;

5. Therefore F is not the centre of the circles ABC, CDE. Conclusion.-Therefore, if one circle, &c. Q. E. D.

PROPOSITION 7.-THEOREM.

If any point be taken in the diameter of a circle, which is not the centre, of all the straight lines which can be drawn from this point to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and, of the others, that which is nearer to the line which passes through the centre, is always greater than one more remote; and from the same point there can be drawn to the circumference two straight lines, and only two, which are equal to one another, one on each side of the diameter.

(References-Prop. I. 4, 20, 23, 24.) Hypothesis.—Let ABCD be a circle, AD its diameter, and E its centre, in which let any point F be taken which is not the centre.

Sequence.-1. Of all the straight lines FB, FC, FG, &c., that can be drawn from F to the circumference, FA, which passes through the centre, shall be the greatest;

2. FD, the other part of the diameter AD, shall be the least;

3. And of the others, FB, the nearer to FA, shall be greater than FC, the more remote; and FC greater than FG.

Construction.-Join BE, CE, GE.

Demonstration.-1. Because any two sides of a triangle are greater than the third side, BE, EF,

А are greater than BF. (I. 20.)

2. But AE is equal to BE; therefore AE, EF, that is, AF is greater than BF.

3. Again, because BE is equal to CE, and EF common to the two triangles BEF, CEF, the two sides BE, EF, are equal to the two CE, EF, each to each;

4. But the angle BEF is greater than the angle CEF;

5. Therefore the base FB is greater than the base FC. (I. 24.)

6. In the same manner it may be shown that FC is greater than FG.

7. Again, because GF, FE, are greater than EG (I. 20), and that EG is equal to ED;

8. Therefore GF, FE, are greater than ED.

B

[ocr errors]

E

K

[ocr errors]

9. Take away the common part FE, and the remainder GF is greater than the remainder FD; (ax. 5.)

10. Therefore FA is the greatest, and FD the least of all the straight lines from F to the circumference; and FB is greater than FC, and FC than FG.

Sequence.-4. Also, there cannot be drawn more than two equal straight lines from the point F to the circumference, one on each side of the diameter.

Construction. At the point E in the straight line EF, make the angle FEH equal to the angle FEG (I. 23), and join FH.

Demonstration.-1. Because EG is equal to EH, and EF common to the two triangles GEF, HEF; the two sides EG, EF, are equal to the two sides EH, EF, each to each;

2. And the angle GEF is equal to the angle HEF; (const.) 3. Therefore the base FG is equal to the base FH. (I. 4.)

4. But, besides FH, no other straight line can be drawn from F to the circumference, equal to FG;

5. For, if it be possible, let FK be equal to FG,

6. Then, because FK is assumed to be equal to FG, and FH is also equal to FG (dem. 3), therefore FH is equal to FK;

7. That is, a line nearer to that which passes through the centre, is equal to a line which is more remote; which is impossible by what has been already shown.

Conclusion. Therefore, if any point, &c. Q. E. D

PROPOSITION 8. -THEOREM.

If any point be taken without a circle, and straight lines be drawn from it to the circumference, one of which passes through the centre; of those which fall on the concave circumference, the greatest is that which passes through the centre, and of the rest, that which is nearer to the one passing through the centre is always greater than one more remote; but of those which fall on the convex circumference, the least is that between the point without the circle and the diameter; and of the rest, that which is nearer to the least is always less than one more remote; and from the same point there can be drawn to the circumference tuo straight lines, and only two, which are equal to one another, one on each side of the line which passes through the centre.

(References—Prop. I. 4, 20, 21, 23, 24; III. 1.) Hypothesis.-Let ABC be a circle, and D any point with

out it, and from D let the straight lines DA, DE, DF, DC, be drawn to the circumference, of which DA passes through the centre.

Sequence.-1. Of the lines which fall on the concave part of the circumference AEFC, the greatest shall be DA, which passes through the centre, and the nearer to it shall be greater than the more remote, viz. DE greater than DF, and DF greater than DC.

2. But of those which fall on the convex circumference GKLH, the least shall be DG between the point D and the diameter AG, and the nearer to it shall be less than the more remote, viz. DK less than DL, and DL less than DH.

Construction.—Take M, the centre of the circle ABC (III. 1), and join ME, MF, MC, MH, ML, MK.

Demonstration.-1. Because any two sides of a triangle are greater than the third side, EM, MD, are greater than ED. (I. 20.) 2. But EM is equal to AM; there

GB fore AM, MD, are greater than ED, that is, AD is greater than ED.

3. Again, because EM is equal to FM, and MD common to the two triangles EMD, FMD; the two sides EM, C MŰ, are equal to the two sides FM, MD, each to each;

4. But the angle EMD is greater than the angle FMD;

5. Therefore the base ED is greater than the base FD. (I. 24.)

6. In like manner it may be shown that FD is greater than CD.

7. Therefore DA is the greatest, and DE greater than DF, and DF greater than DC.

8. Again, because MK, KD, are greater than MD (I. 20), and MK is equal to MG;

9. The remainder KD is greater than the remainder GD, that is, GD is less than KD.

10. And because MK, DK, are drawn to the point K within the triangle MLD from M and D, the extremities of its side

[ocr errors]

M

F

MD;

11. Therefore MK, KD, are less than ML, LD. (I. 21.)

12. But MK is equal to ML; therefore the remainder KD is less than the remainder LD.

13. In like manner it may be shown that LD is less than HD.

14. Therefore DG is the least, and DK less than DL, and DL less than DH.

Sequence.-3. Also, there can be drawn only two equal straight lines from the point D to the circumference, one on each side of the least line.

Construction. At the point M, in the straight line MD, make the angle DMB equal to the angle DMK (I. 23), and join DB.

Demonstration.-1. Because MK is equal to MB, and MD common to the two triangles KMD, BMD; the two sides KM, MD, are equal to the two sides BM, MD, each to each;

2. And the angle DMK is equal to the angle DMB; (const.) 3. Therefore the base DK is equal to the base DB. (I. 4.)

4. But, besides DB, no other straight line can be drawn from D to the circumference, equal to DK.

5. For, if it be possible, let DN be equal to DK.

6. Then, because DN is assumed to be equal to DK, and DB is also equal to DK (dem. 3), therefore DB is equal to DN; (ax. 1.)

7. That is, a line nearer to the least is equal to one which is more remote; which is impossible by what has been already shown.

Conclusion.—Therefore, if any point, &c. Q. E. D.

PROPOSITION 9.-THEOREM.

If a point be taken within a circle, from which there can be drawn more than two equal straight lines to the circumference, that point is the centre of the circle.

(Reference—Prop. III. 7.) Hypothesis.-Let the point D be taken within the circle ABC, from which to the circumference there can be drawn more than two equal straight lines, viz. DA, DB, DC.

Sequence.—The point D shall be the centre of the circle.

Construction. For if D be not the F centre, let E be it; join DE, and produce it to the circumference in F and G.

Demonstration.—1. Then FG is a diameter of the circle ABC. (I. def. 17.)

2. And because in FG, the diameter of the circle ABC, there is taken the point D, not the centre;

3. Therefore DG is the greatest straight line from D to the

DE

[ocr errors]

А

B

« ForrigeFortsett »