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(References-Prop. I. 12, 20; III. def. 5.) Hypothesis I.—Let ABCD be a circle, of which AD is a diameter, and E the centre; and let BC be nearer to the centre than FG.

Sequence.—Then AD shall be greater than any straight line BC, which is not a diameter; and BC shall be greater than FG.

Construction. From the centre E draw EH, EK, perpendiculars to BC, FG (I. 12), and join EB, EC, EF.

Demonstration.-1. Because AE is equal to BE, and ED to EC; 2. Therefore AD is equal to BE, EC;

H 3. But BE, EC, are greater than BC; (I. 20.)

4. Therefore also AD is greater than BC.

5. And because BC is nearer to the centre than FG (hyp.), EH is less than EK. (III. def. 5.)

6. But, as was demonstrated in the preceding proposition, BC is double of BH, and FG double of FK, and the squares on EH, HB, are equal to the squares on EK, KF.

7. But the square on EH is less than the square on EK, because EH is less than EK;

8. Therefore the square on HB is greater than the square on KF, and the straight line BH greater than FK;

9. And therefore BC is greater than FG. Hypothesis II.-Next, let BC be greater than FG. Sequence.—Then BC shall be nearer to the centre than FG, that is, the same construction being made, EH shall be less than EK.

Demonstration.—1. Because BC is greater than FG, BH is greater than FK.

2. But the squares on BH, HE, are equal to the squares on FK, KE;

3. And the square on BH is greater than the square on FK, because BH is greater than FK;

4. Therefore the square on HE is less than the square on KE, and the straight line EH less than EK;

5. And therefore BC is nearer to the centre than FG. (IIL. def. 5.)

Conclusion. Therefore, the diameter, &c. Q. E. D.

PROPOSITION 16. -THEOREM.

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The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle; and a straight line, making an acute angle with the diameter at its extremity, cuts the circle.

(References-Prop. I. 12, 17, 19; III. 2, def. 2.) Hypothesis I.-Let ABC be a circle, of which D is the centre, and AB a diameter, and AE a line drawn from A perpendicular to AB.

Sequence. The straight line AE shall fall without the circle.

Construction.-In AE take any point F; join DF, and let DF meet the circle in C.

Demonstration.--1. Because DAF is a right angle, it is greater than the angle AFD; (I. 17.)

2. Therefore DF greater than DA. (I. 19.)

3. But DA is equal to DC; therefore DF is greater than DC.

4. Therefore the point F is without the circle.

5. In the same manner it may be shown that any other point in AE, except the point A, is without the circle.

6. Therefore AE falls without the circle.

Hypothesis II.-Again, let AG make with the diameter the angle DAG less than a right angle.

Sequence.-The line AG shall fall within the circle, and hence cut it.

Construction.-From D draw DH at right angles to AG, and meeting the circumference in K. (I. 12.)

Demonstration. -1. Because DHA is a right angle, and DAH less than a right angle, being a part of DAE;

2. Therefore the side DH is less than the side DA. (I. 19.)

3. But DK is equal to DA; therefore DH is less than DK.

4. Therefore the point H is within the circle. 5. Therefore the straight line AG cuts the circle. Conclusion.—Therefore, the straight line, &c. Q. E. D. Corollary.-From this it is manifest, that the straight line

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which is drawn at right angles to the diameter of a circle, from the extremity of it, touches the circle (III. def. 2); and that it touches it only in one point, because if it did meet the circle in two points it would fall within it (III. 2). Also it is evident that there can be but one straight line which touches the circle in the same point.

PROPOSITION 17.-PROBLEM.

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To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle.

(References-Prop. I. 4, 11; III. 1, 16, cor.) Given.—First, let the given point A be without the given circle BCD.

Sought. It is required to draw from A a straight line, which shall touch the given circle.

Construction.-l. Find the centre E of the circle (III. 1), and join AE.

2. From the centre E, at the distance EA, describe the circle AFG.

3. From the point D draw DF at right angles to EA (I. 11), and join EBF and AB.

Then AB shall touch the circle BCD.

Proof.-1. Because E is the centre of the circles AFG, BCD, EA is equal to EF, and ED to EB.

2. Therefore the two sides AE, EB, are equal to the two sides FE, ED, each to each;

3. And the angle at E is common to the two triangles AEB, FED;

4. Therefore the base AB is equal to the base FD, and the triangle AEB to the triangle FED, and the other angles to the other angles, each to each, to which the equal sides are opposite; (I. 4.)

5. Therefore the angle ABE is equal to the angle FDE. 6. But the angle FDE is a right angle; (const.) 7. Therefore the angle ABE is a right angle. (ax. 1.) 8. And EB is drawn from the centre. (const.) 9. But the straight line drawn at right angles to a diameter of a circle, from the extremity of it, touches the circle; (III. 16, cor.)

10. Therefore AB touches the circle, and it is drawn from the given point A

11. Next, let the given point be in the circumference of the circle, as the point D.

12. Draw DE to the centre E, and DF at right angles to DE; 13. Then DF touches the circle. (III. 16, cor.)

Conclusion.—Therefore from the given points A and D straight lines, AB and

DF, have been drawn, touching the given circle BCD. Q. E. F.

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PROPOSITION 18.–THEOREM. If a straight line touch a circle, the straight line drawn from the centre to the point of contact, shall be perpendicular to the line touching the circle.

(References-Prop. I. 17, 19; III. 1.) Hypothesis.- Let the straight line DE touch the circle ABC in the point C; take the centre F (III. 1), and draw the straight line FC.

Sequence.-FC shall be perpendicular to DE.

Construction. For, if not, let FG be drawn from the point F perpendicular to DE, meeting the circumference in B.

Demonstration.-1. Because FGC is assumed to be a right angle (hyp.), FCG is an acute angle (I. 17), and to the greater angle the greater side is opposite; (I. 19.)

2. Therefore FC is greater than FG.

3. But FC is equal to FB; therefore FB is greater than FG; the part greater D than the whole, which is impossible.

4. Therefore FG is not perpendicular to DE.

5. In the same manner it may be shown that no other straight line from F is perpendicular to DE, but FC; therefore FČ is perpendicular to DE.

Conclusion.—Therefore, if a straight line, &c. Q. E. D.

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PROPOSITION 19.--THEOREM.

If a straight line touch a circle, and from the point of con tact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line.

(Reference-Prop. III. 18.) Hypothesis.-Let the straight line DE touch the circle ABC in C, and from C let CA be drawn at'right angles to DE.

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Sequence.—The centre of the circle shall be in CA.

Construction.-For, if not, if possible. let F be the centre, and join CF.

Demonstration. - 1. Because DE touches the circle ABC, and FC is drawn from the assumed centre to the point of contact;

2. Therefore FC is perpendicular to DE; (III. 18.)

3. Therefore FCE is a right angle.

4. But the angle ACE is also a right angle; (const.)

5. Therefore the angle FCE is equal to the angle ACE; the less to the greater, which is impossible.

6. Therefore F is not the centre of the circle ABC. 7. In the same manner it may be shown that no other point which is not in CA is the centre; therefore the centre is in CA.

Conclusion.—Therefore, if a straight line, &c. Q. E. D.

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PROPOSITION 20.-THEOREM.

The angle at the centre of a circle is double of the angle at the circumference, upon the same base; that is, upon the same arc.

(References-Prop. I. 5, 32.) Hypothesis.

Let ABC be a circle, and BEC an angle at the centre, and BAC an angle at the circumference, which have the same arc BC for their base.

Sequence. The angle BEC shall be double of the angle BAC.

Case I. First, let the centre E of the circle be within the angle BAC.

Construction.—Join AE, and produce it to the circumference in F.

Demonstration. - . 1. Because EA is equal to EB, the angle EAB is equal to the angle EBA; (I. 5.)

2. Therefore the angles EAB, EBA, are double of the angle EAB.

3. But the angle BEF is equal to the angles EAB, EBA; (I. 32.)

4. Therefore the angle BEF is double of the angle EAB.

5. For the same reason the angle FEC is double of the angle EAC.

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