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(References, Prop. I. 11, 13, 32; III. 19, 22, 31.) Hypothesis.-Let EF be a tangent to the circle ABCD, and BD a chord drawn from the point of contact B, cutting the circle.

Sequence. The angles which BD makes with the tangent EF, shall be equal to the angles in the alternate segments of the circle;

That is, the angle DBF shall be equal to the angle in the segment BAD, and the angle DBE shall be equal to the angle in the segment BCD.

Construction.—1. From the point B draw BA at right angles to EF. (I. 11.)

2. Take any point C in the circumference BD, and join AD, DC, CB.

Demonstration. — 1. Because the straight line EF touches the circle ABCD at the point B (hyp.), and BA is drawn at right angles to the tangent from the point of contact B, (const.)

2. The centre of the circle is in BA. (III. 19.)

3. Therefore the angle ADB, being in a semicircle, is a right angle. (III. 31.)

4. Therefore the other two angles BAD, ABD, are equal to a right angle. (I. 32.)

5. But ABF is also a right angle. (const.)

6. Therefore the angle ABF is equal to the angles BAD, ABD.

7. From each of these equals take away the common angle ABD;

8. Therefore the remaining angle DBF is equal to the remaining angle BAD, which is in the alternate segment of the circle. (ax. 3.)

9. And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD, are together equal to two right angles. (III. 22.)

10. But the angles DBF, DBE, are together equal to two right angles. (I. 13.)

11. Therefore the angles DBF, DBE, are together equal to the angles BAD, BCD.

12. And the angle DBF has been shown equal to the angle BAD;

13. Therefore the remaining angle DBE is equal to the angle BCD, which is in the alternate segment of the circle. (ax. 3.)

Conclusion. Therefore, the angles, &c. 8. E. D.

PROPOSITION 33.-PROBLEM.

Upon a given straight line to describe a segment of a circle, con

taining an angle equal to a given rectilineal angle.

(References—Prop. I. 4, 10, 11, 23, 31; III. 16 cor., 31, 32.)

Given.—Let AB be the given straight line, and C the given rectilineal angle.

Sought.-It is required to describe, on the given straight line AB, a segment of a circle, containing an angle equal to the angle C.

Case I.—Let the angle C be a right angle.

Construction.—1. Bisect AB in C F. (1. 10.)

2. From the centre F, at the distance FB, describe the semicircle

APB АНВ.

Then AHB shall be the segment required.

Proof.—Because AHB is a semicircle, the angle AHB in it is a right angle, and therefore equal to the angle C. (III. 31.)

Case II.—Let C be not a right angle.

Construction.-1. At the point A, in the straight line AB, make the angle BAD equal to the angle C. (I. 23.)

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2. From the point A, draw AE at right angles to AD. (I. 11.) 3. Bisect AB in F. (I. 10.)

4. From the point F draw FG at right angles to AB (I. 11), and join GB.

Then AHB shall be the segment required.

Proof.-1. Because AF is equal to BF (const.), and FG is common to the two triangles AFG, BFG;

2. The two sides AF, FG, are equal to the two sides BF, FG, each to each;

3. And the angle AFG is equal to the angle BFG; (const.) 4. Therefore the base AG is equal to the base BG; (I. 4.)

5. And the circle described from the centre G, at the distance GA, will therefore pass through the point B.

6. Let this circle be described; and let it be AHB.

7. Because from the point A, the extremity of the diameter AE, AD is drawn at right angles to AE; (const.)

8. Therefore AD touches the circle. (III. 16, cor.)

9. Because AB is drawn from the point of contact A, the angle DAB is equal to the angle in the alternate segment AHB. (III. 32.)

10. But the angle DAB is equal to the angle C; (const.)

11. Therefore the angle in the segment AHB is equal to the angle C. (ax. 1.)

Conclusion.—Therefore, on the given straight line AB, the segment AHB of a circle has been described, containing an angle equal to the given angle C. Q. E. F.

PROPOSITION 34.–PROBLEM.

From a given circle to cut off a segment, which shall contain an

angle equal to a given rectilineal angle.

(References, Prop. I. 23; III. 17, 32.) Given.-Let ABC be the given circle, and D the given rectilineal angle.

Sought.—It is required to cut off from the circle ABC a segment that shall contain an angle equal to the angle D.

Construction.-1. Draw the straight line EF touching the circle ABC in the point B; (III. 17.)

2. And at the point B, in the straight line BF, make the angle FBC equal to the angle D. (I. 23.)

Then the segment BAC shall contain an angle equal to the given angle D.

Proof. ---1. Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B; (const.)

2. Therefore the angle FBC is equal to the angle in the alternate segment BAC of the circle. (III. 32.)

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3. But the angle FBC is equal to the angle D; (const.)

4. Therefore the angle in the segment BAC is equal to the angle D. (ax. 1.)

Conclusion.—Therefore, from the given circle ABC, the segment BAC has been cut off, containing an angle equal to the given angle D. Q. E. F.

PROPOSITION 35.-THEOREM.

If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them shall be equal to the rectangle contained by the segments of the other.

(References—Prop. I. 12, 47; II. 5; III. 1, 3.) Hypothesis.-Let the two straight lines AC, BD, cut one another in the point E, within the circle ABCD.

Sequence. The rectangle contained by AE and EC shall be equal to the rectangle contained by BE and ED.

Case I.-Let AC, BD, pass each of them through the centre.

Demonstration. - 1. Because E is the centre, EA, EB, EC, ED, are all equal; (I. def. 15.)

2. Therefore the rectangle AE.EC is equal B to the rectangle BE·ED.

Case II.-Let one of then, BD, pass through the centre, and cut the other, AC, which does not pass through the centre, at right angles, in the point E.

Construction.—Bisect BD in F, then F is the centre of the circle; join AF.

Demonstration. — 1. Because BD, which passes through the centre, cuts AC, which does not pass through the centre, at right angles in E; (hyp.)

2. Therefore AE is equal to EC. (III. 3.)

3. And because BD is cut into two equal parts in the point F, and into two unequal parts in the point E.

4. The rectangle BEED, together with the square on EF, is equal to the square on FB (II. 5); that is, the square on AF.

5. But the square on AF is equal to the squares on AE, EF;

(I. 47.)

6. Therefore the rectangle BEED, together with the square on EF, is equal to the squares on AE, EF. (ax. 1.)

7. Take away the common square on EF;

8. Then the remaining rectangle BEED, is equal to the remaining square on AE; that is, to the rectangle AE EC, since AE is equal to EC.

Case III.—Let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in the point E, but not at right angles.

Construction.-1. Bisect BD in F, then F is the centre of the circle.

2. Join AF, and from F draw FG perpendicular to AC. (I. 12.)

X3 Demonstration. — 1. Then AG is equal to GC; (III. 3.)

2. Therefore the rectangle AEEC, together with the square on EG, is equal to the square on AG. (II. 5.)

3. To each of these equals add the square on GF;

4. Then the rectangle AEEC, together with the squares on EG, GF, is equal to the squares on AG, GF. (ax. 2.)

5. But the squares on EG, GF, are equal to the square on EF;

6. And the squares on AG, GF, are equal to the square on AF. (I. 47.)

7. Therefore the rectangle AE.EC, together with the square on EF, is equal to the square on AF; that is, the square on FB.

8. But the square on FB is equal to the rectangle BEED, together with the square on EF; (II. 5.)

9. Therefore the rectangle AEEC, together with the square on EF, is equal to the rectangle BE: ED, together with the square on ET.

10. Take away the common square on EF;

11. And the remaining rectangle AEEC, is equal to the remaining rectangle BE. ED. (ax. 3.)

Case IV.-Let neither of the straight lines AC, BD, pass through the centre.

Construction.— Take the centre F (III. 1), and through E, the intersection of the lines AC, BD, draw the diameter GEFH.

A Demonstration.-1. Because the rectangle GE EH, is equal, as has been shown, to the rectangle AE·EC, and also to the rectangle BEED;

2. Therefore the rectangle AE. EC is equal to the rectangle BE: ED. (ax. 1.)

Conclusion.—Therefore, if two straight lines, &c. Q. E. D.

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