sequently the other two angles BAD, ABD are equal (5. 2.) to a right angle; but ABF is likewise a right angle; therefore 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 (14. 3.) to two right angles; therefore the angles DBF, DBE being likewise equal 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. XXIV. THEOR.

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 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

B

A

E

D

D.

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 (5. 3.) to one another: and because the straight line BD is cut into two equal parts in the point F, and into two unequal in the point E, the rectangle BE, ED together with the square of EF, is equal (5. 2.) to the square of FB; that is, to the square of FA; but the squares of AE, EF are equal (17.2.) to the square of FA; therefore the rectangle BE, ED, together with the square of EF, is equal to the squares of

away

AE,
EF: take the common square of A
EF, and the remaining rectangle BE, ED
is equal to the remaining square of AE;
that is, to the rectangle AE, EC.

B

F

E

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.

D

Join AF, and from F draw (10. 1.) FG perpendicular to AC; therefore AG is equal (5. 3.) to GC; wherefore the rectangle AE, EC, together with the square of EG, is equal (21. 2.) to the square of AG: to each of these equals 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 (17. 1.) 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 (21. 2.) to the rectangle BE, ED, together with the square of EF: therefore the rectangle AE, EC, together with the square of EF, is equal to the rectangle BE, ED, together with the square of EF: take away the common square of EF, and the remaining rectangle AE, EC is therefore equal to the remaining rectangle BE, EĎ.

A

H

F

E

B

Lastly, Let neither of the straight lines AC, BD pass through the centre: take the centre F, and through E, the intersection of the straight lines AC, DB, draw the diameter GEFH: and because the rectangle AE, EC is equal, as has been shown, to the rectangle GE, EH; and, for the same reason, the rectangle BE, ED is equal to the same rectangle GE, EH; therefore the rectangle AE, EC is equal to the rectangle BE, ED. Wherefore, if two A straight lines, &c. Q. E. D.

F

E

C

Upon a given straight line to describe a segment of a circle, containing an angle equal to a given 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 (8. 1.) AB in F, and from the centre F, at the distance FB, describe the semicircle AHB; therefore the angle AHB in a semicircle is (22. 3.) 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 (14. 1.) the angle BAD equal to the angle C, and from the point A draw (10. 1.) AE

at right angles to AD; bisect (9. 1.) AB in F, and from F draw (9. 1.) 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 (4. 1.) to the base GB; and the circle

C

described from the centre G, with the radius 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 (8. 3.) 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 (23. 3.); 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

D

H

F

A

G

contains an angle equal to the given angle at C. Which was to be done.

PROP. XXVI. 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 given angle D.

A

Draw (9. 3.) the straight line EF touching the circle ABC in the point B, and at the point B, in the straight line BF, make (14. 1.) the angle FBC equal to the angle D; therefore, because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal (23. 3.) to the angle in the alternate segment BAC of the circle: but the angle FBC is

D

E

B

F

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 ABC, containing an angle equal to the given angle D. Which was to be done.

THE

ELEMENT

OF

GEOMETRY.

BOOK IV.

DEFINITIONS.

I.

THE relation of one magnitude to another of the same kind in respect of quantity, may be called a ratio.

II.

Each of the quantities in a ratio may be called a term.

III.

If the ratio of the first to the second is the same with that of the third to the fourth, the quantities may be said to be in geometrical proportion.

IV.

In treating of proportion it is usual to employ a number, line, or letter, to express the quantity, and to consider the quantity without reference to kind.

V.

Proportion consists of two ratios, and each ratio of two terms; there may be, however, only three quantities, the second being repeated.

VI.

And the second term may then be called a mean proportional.

VII.

In proportion the first and third terms may be called antecedents.

VIII.

And the second and fourth consequents.

IX.

The antecedents may be said to be homologous to each other.