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Join AE, and prodụce BA to F; and because BE is equal (I. Def. 15.) to EA, (I. 5.)

1. The angle EAB is equal to EBA; also, because AE is equal to EC,

2. The angle EAC is equal to ECA; wherefore (Ax. 2.)

3. The whole angle BAC is equal to the two angles ABC, ACB: but FAC, the exterior angle of the triangle ABC, is equal (I. 32.) to the two angles ABC, ACB; therefore (Ax. 1.)

4. The angle BAC is equal to the angle FAC; and each of them is therefore (I. Def. 10.) a right angle: wherefore

5. The angle BAC in a semicircle is a right angle. And because the two angles ABC, BAC, of the triangle ABC, are together less (I. 17.) than two right angles, and that BAC is a right angle,

1. ABC must be less than a right angle ; and therefore

2. The angle ABC in a segment ABC, greater than a semicircle, is less than a right angle. And because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal (III. 22.) to two right angles ; therefore

1. The angles ABC, ADC, are equal to two right angles; and ABC is less than a right angle; wherefore the other

2. ADC is greater than a right angle. Besides, it is manifest that the circumference of the greater segment ABC

falls without the right angle CAB, but the circumference of the less segment ADC falls within the right angle CAF.

“And this is all that is meant, when in the Greek text, and the translations from it, the angle of the greater segment is said to be greater, and the angle of the less segment is said to be less, than a right angle.”

COR.—From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal (I. 32.) to the same two; and when the adjacent angles are equal, they are (I. Def. 10.) right angles.

PROP. XXXII.-THEOREM. If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle.

Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn cutting the circle : the angles which BD makes with the touching line EF shall be equal to the angles in the alternate segments of the circle: that is, the angle FBD is equal to the angle which is in the segment DAB, and the angle DBE to the angle in the segment BCD.

From the point B draw (I. 11.) BA at right angles to EF, and take any point C in the circumference BD, and join AD, DC, CB; and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line from the point of contact B, (III. 19.)

1. The centre of the circle ABCD is in BA; therefore (III. 31.)

2. The angle ADB in a semicircle is a right angle, and consequently (I. 32.)

3. The other two angles BAD, ABD, are equal to a right angle: but ABF is likewise a right angle; therefore (Ax. 1.)

4. The angle ABF is equal to the angles BAD, ABD: take from these equals the common angle ABD; therefore (Ax. 3.).

5. The remaining angle DBF is equal to the angle BAD, which is in the alternate segment of the circle; and because ABCD is a quadrilateral figure in a circle, (III. 22.)

6. The opposite angles BAD, BCD, are equal to two right angles: but the angles DBF, DBE, are likewise equal (I. 13.) to two right angles; therefore

7. The angles DBF, DBE, are equal to the angles BAD, BCD; and DBF has been proved equal to BAD; therefore

8. The remaining angle DBE is equal to the angle BCD in the alternate segment of the circle. Wherefore, if a straight line, &c. Q.E.D.

PROP. XXXIII. -PROBLEM. Upon a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle.

Let AB be the given straight line, and the angle C the given recti. lineal angle; it is required to describe upon the given straight line AB a segment of a circle, containing an angle equal to the angle C.

First, let the angle C be a right angle.

FA

Bisect (I. 10.) AB in F, and from the centre F, at the distance FB, describe the semicircle AHB; therefore (III. 31.)

The angle AHB in the semicircle is equal to the right angle at C. But if the angle C be not a right angle:

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At the point A in the straight line AB, make (I. 23.) the angle BAD equal to the angle C, and from the point A draw (I. 11.) AE at right angles to AD; bisect (I. 10.) AB in F, and from F draw (1. 11.) FG at right angles to AB, and join GB.

And because AF is equal to FB, and FG common to the triangles AFG, BFG, the two sides AF, FG, are equal to the two BF, FG, each to each ; and the angle AFG is equal (I. Def. 10.) to the angle BFG; therefore (I. 4.)

1. The base AG is equal to the base GB; and the circle described from the centre G, at the distance GA, shall pass through the point B; let this be the circle AHB: and because from the point A the extremity of the diameter AE, AD is drawn at right angles to AE, therefore (III. 16. Cor.)

2. AD touches the circle AHB; and because AB drawn from the point of contact A cuts the circle, (III. 32.)

3. The angle DAB is equal to the angle in the alternate segment AHB: but the angle DAB is equal (Constr.) to the angle C; therefore also

4. The angle C is equal to the angle in the segment AHB. Wherefore, upon the given straight line AB, the segment AHB of a circle is described which contains an angle equal to the given angle C. Which was to be done.

PROP. XXXIV.-PROBLEM.

To cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle.

Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the angle D.

Draw (III. 17.) the straight line EF touching the circle ABC in the point B; and at the point B in the straight line BF make (I. 23.) the angle FBC equal to the angle D: the segment BAC shall contain an angle equal to the given angle D.

А.

D

B

Because the straight line EF touch the circle ABC, and BC is drawn from the point of contact B, (III. 32.)

1. The angle FBC is equal to the angle in the alternate segment BAC of the circle: but the angle FBC is equal (Constr.) to the angle D; therefore (Ax. 1.)

2. The angle in the segment BAC is equal to the angle D. Wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D. Which was to be done.

PROP. XXXV.---THEOREM.

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

Let the two straight lines AC, BD, within the circle ABCD, cut one another in the point E: the rectangle contained by AE, EC, is equal to the rectangle contained by BE, ED.

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If AC, BD, pass each of them through the centre, so that E is the centre; it is evident that AE, EC, BE, ĚD, being (I. Def. 15.) all equal, likewise

1. The rectangle AE, EC, is equal to the rectangle BE, ED.

But let one of them BD pass through the centre, and cut the other AC, which does not pass through the centre, at right angles, in the point E.

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Then if BD be bisected in F, F is the centre of the circle ABCD. Join AF. And because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E, (III. 3.)

1. AE, EC, are equal to one another: and because the straight line BD is cut into two equal parts in the point F, and into two unequal parts in the point E, (II. 5.)

2. The rectangle BE, ED, together with the square of EF, is equal to the square of FB; that is, to the square of FA: but the squares of AE, EF, are equal (I. 47.) to the square of F4; therefore (Ax. 1.)

3. The rectangle BE, ED, together with the square of EF, is equal to the squares of AE, EF: take away the common square of EF, and (Ax. 3.)

4. The remaining rectangle BE, ED, is equal to the remaining square of AE; that is, to the rectangle AE, EC. Next, let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles.

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Then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw (I. 12.) FG perpendicular to AC; therefore (III. 3.)

1. AG is equal to GC; wherefore (II. 5.)

2. The rectangle AE, EC, together with the square of EG, is equal to the square of AG:

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