The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate

PROP. XXXI. THEOR.

In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

Let ABCD be a circle, of which the diameter is BC, and centre E; and draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC; the angle in the semicircle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle ; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle.

Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal (1. 5.) EBA; also, because a is equal to EC the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB; But FAC, the exterior angle of the triangle ABC, is equal (1.32.) to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a right (1. Def. 10.) angle: Wherefore the angle BAC in a semicircle is a right angle.

And because the two angles ABC, BAC of the triangle ABC are together less (1. 17.) than two right angles, and that BAC is a right angle, ABC must be less than a right angle; and therefore the angle 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 (111. 22.) to two right angles; therefore the angles ABC, ADC are equal to two right angles; and ABC is less than a right angle; wherefore the other 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 to the same two; and when the adjacent angles are equal, they are right angles.

PROP. XXXII. THEOR.

If a straight line touches 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, the centre of the circle is (II. 19.) in BA; therefore the angle ADB in a semicircle is a right (I. 31.) angle, and consequently the other two angles BAD, ABD are equal (1. 32.) to a right angle: But ABF is likewise a right angle; there- E

fore the angle ABF is equal to the angles BAD, ABD: Take from these equals the common angle ABD; therefore 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, the opposite angles BAD, BCD are equal (m. 22.) to two right angles; therefore the angles DBF, DBE, being likewise equal (1. 13.) to two right angles, are equal to the angles BAD, BCD; and DBF has been proved equal to BAD: Therefore 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. PROB.

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 at c the given rectilineal 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 at c be a right angle, and bisect (1. 10.) AB in F, and C from the centre F, at the distance FB, describe the semicircle AHB; therefore the angle AHB in a semicircle is (шI. 31.) A equal to the right angle at c.

But, if the angle c be not a right angle, at the point a, in the straight line AB, make (1. 23.) the angle BAD equal to the angle C, and from the point A draw (I. 11.) AE at right angles to AD; bisect (1. 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; and the angle AFG is equal to the angle BFG; therefore the base AG is equal (1. 4.) 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 AD (III. 16. Cor.) touches the circle;

and because AB drawn from the point of contact a cuts the circle, the angle DAB is equal to the angle in the alternate segment AHB (111. 32.). But the angle DAB is equal to the angle C, therefore also 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 at c. Which was to be done.

PROP. XXXIV. PROB.

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 (1. 23.) the angle FBC equal to the angle D. Then, because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal (II. 32.) to the angle in the alternate segment BAC of the circle:

But the angle FBC is equal to the angle D; therefore the angle in the segment BAC is equal to the angle D: Wherefore the segment BAC is cut off from the given circle A BC containing an angle equal to the given angle D: Which was to be done.

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.

If AC, BD pass each of them through the centre, so that E is the centre; it is evident, that AE, EC, BE, ED, being all equal, the rectangle AE, EC is likewise equal to the B 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: 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, AE, EC are equal (111. 3.) 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, the rectangle BE, ED together with the square of EF, is equal (II. 5.) to the square of FB; that is, to the square of FA; but the squares of AE, EF are equal (1. 47.) to the square of FA; therefore 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 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: Then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw (1. 12.) FG perpendicular to AC; therefore AG is equal (111. 3.) to GC; wherefore the rectangle AE, EC, together with the square of EG, is equal (II. 5.) to the square of AG: To each of these squares add the square of GF; therefore the rectangle AE, EC, together with the squares of EG, GF, is equal to the squares of AG, GF: But the squares of EG, GF are equal to the square of EF; and the squares of AG, GF are equal to the square of AF: Therefore the rectangle AE, EC, together with the square of EF, is equal to the square of AF; that is, to the

square of FB: But the square of FB is equal (п. 5.) to the

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The first three books of Euclid's Elements of geometry, with theorems and ...