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scribed upon the right lines AB, BC, CD, DA, the triangles Amв, вnc, CrD, DSA will each of them be half the parallelogram in which it ftands (I. 32.)

But every fegment is less than the parallelogram which circumfcribes it; and therefore each of the triangles AmB, Bnc, crD, DSA is greater than half the fegment of the circle which contains it.

And, if each of the arcs Am, mв, &c. be again divided into two equal parts, and right lines be drawn to the points of bisection, the triangles thus formed, may in like manner, be fhewn to be greater than half the fegments which contain them; and fo on continually.

Since, therefore, the circle ABCD is greater than the space s, and from the former there has been taken more than its half, and from the remainder more than its half, &c. there will at length remain segments which, taken together, fhall be lefs than the excefs of the circle ABCD above the space s (VIII. 1.), as was to be shewn.

PRO P. IV. THEOREM.

A polygon may be circumfcribed about a circle that shall differ from it by less than any affigned magnitude whatever,

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Let ABCD be the circle, and s any given magnitude whatever; then may a polygon be circumfcribed about

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the circle ABCD, that fhall differ from it by lefs than the magnitude s.

For let the circle ABCD be circumfcribed by the fquare IFGH (IV. 7.), and bifect the arcs AB, BC, CD, DA with the lines OE, OF, OG and OH; and to the points of bifection draw the tangents kl, mn, pr, st (III. ro.)

Then fince kl is a tangent to the circle, and oɛ is drawn from the centre through the point of contact, the angle Exk is a right angle (III. 12.), and Ek will be greater than kx (I. 17.) or its equal ka.

But triangles of the fame altitude are to each other as their bases (VI. 1.); whence the base Ek being greater than the base kA, the triangle Exk will alfo be greater than the triangle kxa.

And because the triangle Exk is greater than the triangle kxA, it will also be greater than half the curvelineal space EXA and the fame may be fhewn of any other triangle and the curvelineal space to which it belongs.

In like manner, if the arcs Ax, xв, &c. be again bifected, and tangents be drawn to the points of bifection, the triangles thus formed will be greater than half the curvelineal spaces to which they belong.

Since, therefore, the excess of the fquare above the circle is greater than the magnitude s, and from the former there has been taken more than its half, and from the remainder more than its half, and fo on, there will at length remain spaces, which, taken together, fhall be lefs than the magnitude s (VIII. 1.), as was to be shewn.

PROP. V. THEOREMS.

Circles are to each other as the squares of their diameters.

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Let ABCD, EFGH be two circles, and BD, FH their diameters: then will the fquare of BD be to the square of FH as the circle ABCD is to the circle EFGH.

For, if they have not this ratio, the fquare of BD will be to the fquare of FH, as the circle ABCD is to fome space either lefs or greater than the circle EFGH.

First, let it be to a space ST less than the circle EFGH; and infcribe the two fimilar polygons AROPQ, EKLMN fo that the circle EFGH may exceed the latter by less than it exceeds the space ST (VIII. 3.)

Then, fince the circle EFGH exceeds the polygon EKLMN by less than it exceeds the space ST, the polygon EKLMN will be greater than the space ST.

And, becaufe fimilar polygons, infcribed in circles, are to each other as the fquares of their diameters (VIII. 2.), the fquare of BD will be to the fquare of FH as the polygon AROPO is to the polygon EKLMN.

But the fquare of BD is also to the fquare of FH as the circle ABCD is to the space ST (by Conft.); whence the circle ABCD will be to the fpace ST, as the polygon AROPQ is to the polygon EKLMN.

The circle ABCD, therefore, being greater than the polygon AROPQ, which is contained in it, the space ST will also be greater than the polygon EKLMN.

It is, therefore, lefs and greater at the fame time, which is impoffible; confequently the fquare of BD is not to thẹ fquare of FH as the circle ABCD is to any space less than the circle EFGH.

And, in the fame manner, it may be demonftrated, that the fquare of FH is not to the fquare of BD as the circle EFGH is to any space less than the circle ABCD.

Nor, is the fquare of BD to the fquare of FH as the circle ABCD is to a space greater than the circle EFGH. For, if it be poffible, let it be so to the space sx, which is greater than the circle EFGH.

Then, fince the fquare of BD is to the fquare of FH as the circle ABCD is to the space sx, therefore, also, inverfely, the fquare of FH is to the fquare of BD as the fpace sx is to the circle ABCD (V. 7.)

But the space sx is greater than the circle EFGH (by Hyp.); whence the space sx is to the circle ABCD as the circle EFGH is to fome space less than the circle ABCD (V. 14.)

The fquare of FH is, therefore, to the fquare of BD as the circle EFGH is to a space lefs than the circle ABCD (V. 11.), which has been fhewn to be impoffible.

Since, therefore, the fquare of BD is not to the square of FH as the circle ABCD is to any space either lefs or greater than the circle EFGH, the fquare of BD muft be to the fquare of FH as the circle ABCD is to the circle Q. E. D. COR. 1. Circles are to each other as the fquares of their radii; thefe being half the diameters.

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COR. 2. If the radii or diameters of three circles be respectively equal to the three fides of a right angled triangle, that whofe radius or diameter is the hypothenuse will be equal to the other two taken together (II. 14.)

PROP. VI. THEOREM.

Every circle is equal to the rectangle of its radius, and a right line equal to half its circumference.

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Let kmps be a circle, and ov a rectangle contained under the radius ok and a right line ow equal to half the circumference; then will the circle kmps be equal to the rectangle ov.

For if it be not, it must be either greater or less.

Let it be greater; and let the rectangle oz be equal to the circle kmps; and infcribe a polygon In rt in the circle kmps that fhall differ from it by less than the magnitude w2 (VIII. 3.)

Then fince the triangle kot is equal to half a rectangle under the base kt and the perpendicular ox (I. 32.), the whole polygon will be equal to half a rectangle under its perimeter and the perpendicular ox.

And because ow is greater than half the perimeter of any polygon that can be infcribed in the circle kmps

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