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the sides AE and EF; and FH will be equal to their dife ference.

But the rectangle under the sum and difference of the two sides 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 same manner, it may be proved, that the rectangle of IF, FH, is equal to the rectangle of DF, FC: consequently the rectangle of AF, FB is also equal to the rectangle of DF, FC.

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

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PRO P. 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.

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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, co 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, since Aec is a triangle, and the perpendicular 2G divides the chord Ad into two equal parts (III. 3.), the line DC will be equal to the difference of the segments AG, GC.

And, because e is the centre of the circle, and AE is equal to Eh, or El, the line hc will be equal to the fum of the sides AE, EC, and ic will be equal to their difference.

But the rectangle under the fum and difference of the two sides of any triangle, is equal to the rectangle under the base and the difference of the segments of the base (Cor. II. 16.); whence the rectangle of rc, ci is equal to the rectangle of AC, CD.

And, in the same manner, it may be proved, that the rectangle of hc, ci is also 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.

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PRO P. XXIX. THEOREM.

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 square of the other.

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Let CB, CA de 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 square of ca.

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

Then, since 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 sum of EA, EC, and CD will be equal to their difference.

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

But the difference of the squares of CE, EA is equal to the square 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; consequently the rectangle of CB, CF, will also be equal to the square of ca.

R. E, D.

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

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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 square 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.); also join FD, FA, FC.

Then, since 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 also equal to the square of ac (by Hyp.); whence the square of ac is equal to the square of ad, or ac equal to AD (II. 3.)

And,

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