together less than two right angles (I. 17), and BAC has been proved to be a right angle. Therefore A B C is less than a right angle. Wherefore the angle in a segment ABC greater than a semicircle, is less than a right angle. Because ABCD is a quadrilateral figure inscribed in a circle, any two of its opposite angles are equal (III. 22) to two right angles. Therefore the augies A B C and ADC, are equal to two right angles. But ABC has been proved to be less than a right angle. Therefore the angle ADC is greater than a right angle. Wherefore the angle in a segment ADC less than a semicircle is greater than a right angle. Therefore, in a circle, &c. Q. E. D.

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 (I. Def. 10) they are right angles.

Dr. Thomson, in his edition, simplifies the demonstration of this proposition, by the
application of Prop. XX., after the following manner:-The angle BAC at
the circumference is half the two angles at the centre formed by producing A E.
But these angles are equal to two right angles. Therefore BAC is a right angle;
and it is the angle in a semicircle BA C. The angle BAD is greater than the
angle BAC (I. Ax. 9). Therefore the angle BAD is greater than a right angle;
and it is the angle in a segment BAD, less than a semicircle.
In like manner,

by drawing a straight line from the point A, within the angle B AC, it may be
shown that the angle in a segment greater than a semicircle is less than a right
angle.

Exercise. To draw a perpendicular to a straight line from one of its extremities, without producing it, by means of the first part of this proposition.

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 which this straight line makes with the tangent are equal to the angles in the alternate segments of the circle.

Let the straight line EF touch the circle ABCD at the point B; and from the point B, let the straight line BD be drawn, cutting the circle, and dividing it into the two segments DCB and DAB, of which DCB is less, and DAB greater than a semicircle. The angles which BD makes with the tangent EF are equal to the angles in the alternate segments of the circle; that is, the angle DBF is equal to the angle in the segment DAB, and the angle DBE to the angle in the segment DCB. From the point B, draw BA at right angles (I. 11)

to EF, take any point C in the arc DCB, and join AD, DC, and CB.

Because the straight line EF touches the circumference of the circle ABCD at the point B, and BA is drawn at right angles to the tangent from the point of contact B, the centre of the circle is (III. 19) in

F

BA. Therefore the angle ADB in a semicircle (III.31) E B is a right angle. Because the other two angles BAD and ABD, in the triangle ADB, are equal to (I. 32) a right angle, and ABF is (Const.) a right angle. Therefore the angle ABF is equal to the two angles BAD and ABD (I. Ax. 1). From these equals take

away the common angle ABD. Therefore the remaining angle DBF is equal to the remaining angle BAD (I. Ax. 3), which is in the alternate segment of the circle. Because ABCD is a quadrilateral figure in a circle, the opposite angles BAD and BCD are equal (III. 22) to two right angles. But the two angles DBF and DBE are equal (I. 13) to two right angles. Therefore the two angles DBF and DBE are equal (I. Ax. 1) to the two angles BAD and BCD. But the angle DBF has been proved equal to the angle BAD. Therefore the remaining angle DBE is equal (I. Ax. 2) to the remaining angle B CD, which is in the alternate segment of the circle. Wherefore, if a straight line, &c. Q. E. D.

In Dr. Thomson's edition, the case, in which the straight line cutting the circle passes through the centre, is considered; this is so evident, that it appears to be an unnecessary addition.

Corollary 1.-If the angles which any straight line meeting a circle makes with a chord, be equal to the angles in the alternate segments, that straight line touches the circle.

Corollary 2.-Tangents to a circle at the extremities of the same chord are equal, and make equal angles with it on the same side.

Corollary 3.-The chord which joins the points of contact of parallel tangents is a diameter.

PROP. XXXIII. PROBLEM.

Upon a given straight line to describe a segment of a circle, which shall contain an angle equal to a given rectilineal angle.

Let AB be the given straight line, and C the given rectilineal angle. It is required to describe upon the given straight line AB, a segment of a circle, which shall contain 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.

H

Because

the segment AHB is a semicircle, the angle AHB is

an angle (III. 31), and therefore it is equal to the given angle C. Next, let the angle C be an oblique angle.

At the point A, in the straight line AB, make the angle BAD equal

to the angle C (I. 23); and from the point A, draw A E at right angles (I. 11) to AD. Bisect A B in F (I. 10); and from F, draw FG at right angles (I. 11) to AB. Join GB.

Because AF is equal to FB, and FG common to the triangles AFG and BFG, the two sides AF and FG are equal to the two sides BF and FG, each to each. But the angle AFG is equal (I. Def. 10) to the

From a gires corde to cut of a sermen, rich shaă comain on cage equal to a garen restitued capit.

Le ABC be the given érds, and Ɔ the given rectilized angle. It is repaired to cut off from the circle AEC a segment that shall contain an ango na to the given angle D.

Daw the staiga Be EF touching the dre ABC in the point B (IL 17), and at the pin B, in the straight line BF, make the anger FBC equal L 25) to the angle D. The wigment BAC contains an angle equal to the given angle 1),

Because the straight line EF touches the Gros ABC, and BC is drawn from the point of

ontact B, the angle FBC is equal (IIL 32) to the angle in the alternate wegment BAC of the circle. But the angle FBC is equal (Const.) to the angle D. Therefore the angle in the segment BAC is equal (I. Az. 1) to the angle D. Wherefore, from the given circle A B C, the segment BAC is cut off, containing an angle equal to the given angle D. QE. F.

Beercise-Through a given point to draw a straight line that shall cut off, from a given circle, a segment containing a given angle.

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

In the circle ABCD let the two straight lines AC and BD, cut one another in the point E. The rectangle contained by AE and EC is equal to the rectangle contained by B E and E D.

First, let AC and BD pass each through the centre E. Because the segments AE, EC, BE and ED are all equal (I. Def. 15). Therefore the rectangle AE.E C, is equal to the rectangle

BD.ED.

Secondly, let the one BD pass through the centre, and cut the other A C, which does not pass through the centre, at right angles, in the point E.

...D

F

Find F, the centre of the circle ABCD, and join AF. Because BD passing through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E (III. 3), A E is equal to E C. Because the straight line BD is cut into two equal parts at the point F, and into two unequal parts at 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, because FA is equal to F B. But the squares of A E and EF, are equal (I. 47) to the square of F A. Therefore the rectangle BE.ED together with the square of EF, is equal (I. Ax. 1) to the squares of AE and EF. From these equals take away the common square of EF. Therefore the remaining rectangle BE.ED is equal (I. Ax. 3) to the remaining square of AE; that is, to the rectangle A E.EC.

B

E

C

Thirdly, let BD, passing through the centre, cut the other A C, which does not pass through the centre, at E, but not at right angles. Find F the centre of the circle, join A F, and from F draw FG perpendicular (I. 12) to A C. Because A G is equal (III. 3) to GC, the rectangle AE.EC, together with the square of EG, is equal (II. 5) to the square of AG. To each of these equals, add the square of GF. Therefore the rectangle AE.E C, together with the squares of E G and GF, is equal (I. Ax. 2) to the squares of AG and G F. But the squares of EG and GF, are equal (I. 47) to the square of EF. And the squares of AG and GF are equal to the square of AF. Therefore the rectangle A E.E C, together with the square of EF, is equal to the square of A F; that is, to the square of FB. But the square of FB is equal (II. 5) to the rectangle BE.ED, together with the square of EF. Therefore the rectangle AE.EC, together with the square of E F, is equal (I. Ax. 1) to the rectangle B E.ED, together with the square of ÉF. From these equals take away the common square of E F. Therefore the remaining rectangle A E.E C, is equal (Ax. 3) to the remaining rectangle

BE.ED.

Lastly, let neither of the straight lines A C and BD cutting each other, pass through the centre.

Find the centre F (III. 1), and through E, the point of intersection of the straight lines AC and DB, draw the diameter G H.

Because the rectangle A E.EC is equal, by the preceding case, to the rectangle GE.EH. And the rectangle BE.ED is equal to the same rectangle

B

GE.E H. Therefore the rectangle A E.EC is equal (I. Ax. 1) to the rectangle B E.ED. Wherefore, if two straight lines, &c. Q. É. D.

The second case of this proposition is included in the demonstration of Prop. XIV, Book II. A demonstration including all the cases may be derived from Props. IV. and XVI., of Book VI.

Corollary.-If the rectangles contained by the segments of the diagonals of any quadrilateral figure, be equal, a circle may be described about it.

PROP. XXXVI. THEOREM.

If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, is equal to the square of the tangent.

Let D be any point without the circle ABC, from which the two straight lines DA and DB are drawn, of which DA cuts the circle and DB touches it. The rectangle AD.DC is equal to the square of DB.

First, let DA pass through the centre E.

Join EB, and (III. 18) EBD is a right angle.

Because the straight line AC is bisected in E, and produced to D, the rectangle AD.DC, together with the square of EC, is equal (II. 6) to the square of ED. But CE is equal to EB. Therefore the rectangle AD.DC, together with the square of EB, is equal to the square of ED. But the square of ED is equal (I. 47) to the squares of EB and BD. Therefore the rectangle AD.DC, together with the square of EB, is equal (I. Ax. 1) to the squares of EB and BD. From these equals, take away the common square of EB. Therefore the remaining rectangle AD.DC is equal (Ax. 3) to the square of the tangent DB. Next let DA not pass through the centre of the circle ABC.

A

Find E the centre of the circle (III. 1), draw EF perpendicular to

AC (I. 12) and join EB, EC, and ED.

B

D

Because the straight line EF, passing through the centre, cuts the straight line AC, which does not pass through the centre, at right angles, it bisects AC (III. 3). Therefore AF is equal to FC. Because the straight line AC is bisected in F, and produced to D, the rectangle AD.DC together with the square of FC, is equal (II. 6) to the square of FD. To each of these equals, add the square of FE. Therefore the rectangle AD.DC, together with the squares of CF and FE is equal (I. Ax. 2) to the squares of DF and FE. But the square of ED is equal (I. 47) to the squares of DF and FE. Also, the square of EC is equal to the squares of CF and FE. Therefore the rectangle AD.DC, together with the square of EC is equal (I. Ax. 1) to the square of ED. But CE is equal to EB. Therefore the rectangle AD.DC, together with the square of E B, is equal to the square of ED. But the squares of EB and BD are equal (1. 47) to the square of ED. Therefore the rectangle AD.DC, together with the square of EB, is equal to