From the point B draw BA at right angles to EF; (1. 11) and take any point C in the circumference BD, and join AD, DC, CB.

DEMONSTRATION

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 ABCD is in BA; (III. 19)

therefore the angle ADB in a semicircle, is a right angle, (III. 31)

and consequently the other two angles BAD, ABD, are equal to a right angle; (1.32)

but ABF is likewise a right angle; (constr.)

therefore the angle ABF is equal to the angles BAD, ABD; (ax. 1) 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. (ax. 3.)

And because ABCD is a quadrilateral figure in a circle,

the opposite angles BAD, BCD, are equal to two right angles; (III. 22)

but the angles DBF, DBE, are likewise equal to two right angles; (1. 13)

therefore the angles DBF, DBE, 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, which is in the alternate segment of the circle. (ax. 3.)

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.

(References-Prop. 1. 4, 10, 11, 23, 31; III. 16 cor., 31, 32.)

Let AB be the given straight line, and the angle 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 C be a right angle.

Bisect AB in F, (1. 10) and from the centre F, at the distance FB, describe the semicircle AHB.

Then AHB shall be the segment required.

DEMONSTRATION

Because AHB is a semicircle,

therefore the angle AHB in it is a right angle, and therefore equal to C; (III. 31)

wherefore the angle in the segment AHB is equal to the given angle C.

Secondly, let C be not a right angle.

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

and from the point A draw AE at right angles to AD; (1. 11)

bisect AB in F, (1. 10) and from F draw FG at right angles to AB, (1. 11) and join GB.

Then AHB is the segment required.

DEMONSTRATION

In the triangles AFG, BFG,

because AF is equal to FB, and FG common to both,

the two sides AF, FG, are equal to the two BF, FG, each to each; and the angle AFG is equal to the angle BFG; (1. def. 10)

therefore the base AG is equal to the base GB; (1.4)

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 touches the circle AHB; (ш. 16, cor.)

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; (III. 32)

but the angle DAB is equal to the angle C; (constr.)

therefore 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 Q. E. F.

C.

PROP. XXXIV.-PROBLEM.

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

(References-Prop. I. 23; III. 17, 32.)

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 the straight line EF touching the circle ABC in the point B;

(III. 17)

G

and at the point B, in the straight line BF make the angle FBC equal to the angle D. (1. 23.)

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

DEMONSTRATION

Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact, B,

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

but the angle FBC is equal to the angle D; (constr.)

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

Wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D.

Q. E. F.

PROP. XXXV.—THEOREM.

If two straight lines within a circle cut one another;

then the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.

(References Prop. I. 12, 47; II. 5; III. 1, 3.)

Let the two straight lines AC, BD, within the circle ABCD, cut one another in the point E.

Then the rectangle contained by AE, EC, shall be equal to the rectangle contained by BE, ED.

A

E

B

First, let AC, BD, pass each of them through the centre.

DEMONSTRATION

Then because E is the centre,

therefore AE, EC, BE, ED, are all equal to one another, (1. def. 15)

and therefore the rectangle AE, EC, is equal to the rectangle BE, ED.

Secondly, 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.

D

C

B

CONSTRUCTION

Then if BD be bisected in F, F is the centre of the circle ABCD. Join AF.

DEMONSTRATION

Because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E,

therefore AE, EC, are equal to one another, (III. 3)

and because the straight line BD is cut into two equal parts in the point F, and into unequal parts in the point E,

the rectangle BE, ED, together with the square of EF, is equal to the square of FB; (II. 5) that is to the square of FA;

but the squares of AE, EF, are equal to the square of FA; (1. 47) therefore the rectangle BE, ED, together with the square of EF, is equal to the squares of AE, EF; (ax. 1)

take away the common square of EF, and the remainders are equal, that is,

the rectangle BE, ED, is equal to the square of AE, that is to the rectangle AE, EC.

Thirdly, 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.

CONSTRUCTION

Then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw FG perpendicular to AC. (1. 12.)

DEMONSTRATION

Then AG is equal to GC; (III. 3)