of the mean is in fact the rectangle under the two equal means of the three continued proportionals regarded as four proportionals with equal means.

fig. 124.

(266.) If two chords intersect each other in a circle, the rectangle under the segments of the one will be equal to the rectangle under the segments of the other. Let A B and C D be two such chords intersecting at O; draw the lines BC and DA; the angles ADC and ABC standing on the same arc of the circle are equal (110.), and the angles at O in the two triangles are also equal (20.); therefore the triangles ADO A and CBO are mutually equiangular (57.), and are therefore similar. Hence (249.)

D

AO: DO=C0: BO.

B

The rectangle under the means will then be equal to the rectangle under the extremes; that is, the rectangle under AO and BO is equal to the rectangle under DO and CO.

It is evident that the same will be true for any number of chords intersecting in the same point; the rectangle under the segments of each of them will have the same magnitude.

fig. 125.

D

(267.) If AB (fig. 125.) be the diameter of a circle, a perpendicular to it from any point C, meeting the circle at D, will be a mean proportional between the segments AC and CB of the diameter.

For it has been already proved (116.) that if D E be perpendicular to the diameter AB, it will be bisected by the

E

B

is equal to the rectangle under D C and C E, and DC is equal to C E, the rectangle under AC and C B will be equal to the square of DC; therefore DC will be a mean proportional between AC and CB. (265.) (268.) If from the same point P (fig. 126.) on the

circumference of a circle, a tangent PT, and a chord PC, be drawn, the angle CPT, formed by these lines, will be equal to an angle contained in the segment of the circle PAC which lies on the other side of the chord.

For from P, through the centre O, draw the diameter POA, and draw A C; the angle P A C is equal to all the other angles, such as P B C, in the same segment; and it is therefore necessary only to prove that the angle CPT is equal to the angle PAC.

The angle A PT is a right angle (83.), and therefore APC and CPT are together equal to 90°; also the angle ACP is a right angle (112.), and therefore the angles CAP and CPA are together equal to a right angle (52.). Since the angle CAP, together with CPA, makes up 90°, and also CPT, together with the same angle CPA, is 90°; the angle CPT is equal to the angle C A P, and therefore equal to any angle in the same segment (110).

(269.) If from the same point P (fig. 127.) outside a circle a tangent PT and a secant PAS be drawn, the

square of the tangent PT will be equal to the rectangle under PA and PS.

For let TA and TS be drawn: the angle PTA will be equal to the angle S (268.), and the angle P is common to the two triangles PAT and PTS; therefore the triangles are equiangular (57.), and are therefore similar; therefore their corresponding sides are proportional (249.): hence

PA: PTPT: PS.

That is, PT is a mean proportional between PA and PS, and therefore the square of PT is equal to the rectangle under PA and PS.

(270.) Since this will be equally true of all secants drawn from the same point P, it follows that the rectangle under the corresponding lines for each secant are equal. Thus, if PA ́S ́ be drawn, the rectangle under PA' and PS' may in the same manner be proved to be equal to the square of PT, and is therefore equal to the rectangle under PA and PS, and the same will be true of all secants drawn from the same point P.

(271.) To find by geometrical construction a fourth proportional to three lines, is equivalent to the problem to construct upon a given right line a rectangle equal to a given rectangle ; for the fourth proportional will be the height of a rectangle formed on the first of the three given lines whose area is equal to the rectangle under the second and third. Or the question may be stated thus, The means and one extreme of four proportionals are given, and the other extreme is sought.

rectangle under the extremes, the problem is to construct upon the given extreme a rectangle whose area shall be equal to the rectangle under the means.

The principles of geometry which have been already explained present many methods of doing this.

I. Let a (fig. 128.) be the given extreme, and b and c fig. 128.

the given means. Draw two lines MN and MO equal respectively to a and b, and draw NO so as to form a triangle; also draw M'N' equal to c, and on M'N' construct a triangle having its angles equal to those of MNO; the two triangles being respectively equiangular, will be similar, and therefore their sides will be proportional: hence

MN: MOM'N': M'O',

or a:bc: M'O';

M'O' is therefore the fourth proportional which is sought, and the rectangle under M'O' and a will be equal to the rectangle under b and c.

It will be perceived that the spirit of this solution consists in making the given extreme and one of the means two sides of a triangle, and in constructing a similar triangle of which the other mean and the sought extreme shall be corresponding sides. Although the other varieties of solution for this problem are apparently different from the present, yet if carefully considered they will be found to be identical with it; the only difference being in the method by which the two similar triangles are constructed.

II. The same problem may be solved otherwise, thus:

Let the given means or lines equal to them be placed so as to form one straight line, so that A B (fig. 129.)

shall be equal to b, and B C to c; from B draw BD equal to a, and making any angle with AC; through the points A, D, C, describe a circle (121.); produce the line D B until it meet the circle at the opposite side E; BE will then be the fourth proportional sought.

For the rectangle under A B and B C is equal to the rectangle under D B and B E (266.). Hence (262.) DB: AB=BC: BE;

that is,

abc: BE.

III. The problem may also be solved thus: : Draw any two lines A X (fig. 130.) and A Y forming any angle with each other; take upon AX from A two parts AB and A C, equal to the given extreme a and to one of the means b; on the other line A Y take a part A D from A equal to the other mean c; draw a line joining B and D, and from C draw another line parallel to BD which will meet AY at E; AE will then be the fourth proportional sought.

K