the fides AE and EF; and FH will be equal to their dif.. ference.

But the rectangle under the fum and difference of the two fides of any triangle, is equal to the rectangle under the base and the difference of the fegments of the base (Cor. II. 16.); whence the rectangle of IF, FH is equal to the rectangle of AF, FB,

And, in the fame manner, it may be proved, that the rectangle of IF, FH, is equal to the rectangle of DF, FC: confequently the rectangle of AF, FB is also equal to the rectangle of DF, FC.

Q.E.D. SCHOLIUM. When the two lines intersect each other in the centre of the circle, the rectangles of their fegments will manifeftly be equal, because the segments themselves are all equal,

PROP.

PROP. XXVIII.

THEOREM.

If two right lines be drawn from any point without a circle, to the opposite part of the circumference, the rectangle of the whole and external part of the one, will be equal to the rectangle of the whole and external part of the other.

G

H

Let ADFB be a circle, and AC, BC any two right lines, drawn from the point c, to the opposite part of the circumference; then will the rectangle of AC, CD be equal to the rectangle of BC, CF.

For, through the centre E, and the point c, draw the right line CH; and, from the point E, draw EG at right angles to AC (I. 12.) and join AE.

Then, fince AEC is a triangle, and the perpendicular EG divides the chord AD into two equal parts (III. 3.), the line DC will be equal to the difference of the fegments

AG, GC.

And, because E is the centre of the circle, and AE is equal to EH, or E1, the line HC will be equal to the fum of the fides AE, EC, and IC will be equal to their difference.

But the rectangle under the fum and difference of the two fides of any triangle, is equal to the rectangle under the base and the difference of the fegments of the base (Cor. II. 16.); whence the rectangle of HC, CI is equal the rectangle of AĆ, CD.

And, in the fame manner, it may be proved, that the rectangle of HC, CI is alfo equal to the rectangle of CB, CF consequently the rectangle of AC, CD will be equal to the rectangle of CB, CF. Q.E.D.

If two right lines be drawn from any point without a circle, the one to cut it, and the other to touch it; the rectangle of the whole and external part of the one, will be equal to the fquare of the other.

Let CB, CA be any two right lines drawn from the point c, the one to cut the circle ADBG, and the other to touch it; then will the rectangle of CB, CF be equal to the fquare of CA.

For,

For, find E, the centre of the circle (III. 1.), and through the points E, C draw the line CEG; and join EA:

Then, fince AC is a tangent to the circle, and EA is a line drawn from the centre to the point of contact, the angle CAE is a right angle (III. 12,)

And, because EA is equal to EG, or ED, the line CG will be equal to the fum of EA, EC, and CD will be equal to their difference.

Since, therefore, the rectangle under the fum and difference of any two lines is equal to the difference of their fquares (II. 13.), the rectangle of CG, CD will be equal to the difference of the squares of CE, EA.

But the difference of the fquares of CE, EA is equal to the fquare of CA (Cor. II. 14.); therefore the rectangle of CG, CD will also be equal to the square of CA.

And it has been shewn, in the last propofition, that the rectangle of CG, CD is equal to the rectangle of CB, CF; confequently the rectangle of CB, CF, will also be equal to the fquare of ca.

Q.E.D.

PROP. XXX. THEOREM.

If two right lines be drawn from a point without a circle, the one to cut it, and the other to meet it; and if the rectangle of the whole and external part of the one be equal to the fquare of the other, the latter will be a tangent to the circle.

Let AB, AC be two right lines, drawn from any point A, without the circle CBD, the one to cut it, and the other to meet it: then, if the rectangle of AB, AE be equal to the fquare of AC, the line AC will be a tangent to the circle.

For, let F be the centre; and from the point A draw AD to touch the circle at D (III. 10.); alfo join FD FA, FC.

Then, fince AD is a tangent to the circle, and AEB cuts it, the rectangle of AB, AE is equal to the square of AD (III. 29.)

But the rectangle of AB, AE is alfo equal to the fquare of AC (by Hyp.); whence the fquare of AC is equal to the square of AD, or AC equal to AD (II. 3.)