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cor.) the diagonal AB is to the corresponding diagonal CD, as the perimeter AEFBGH to the perimeter CIKDLM. But this proportion must subsist, whatever be the number of chords inscribed in either circumference. Insert a dodecagon in each circle between the hexagon and the circumference, and its perimeter will evidently ap

proach nearer to the length of that circumference. Proceeding thus, by repeated duplications,—the perimeters of the series of polygons that arise in succession, will continually approximate to the curvilineal boundary, which forms their ultimate limit. Wherefore this extreme term, or the circumference AEFBGH, is to the circumference CIKDLM, as the diameter AB to the diameter CD. Again, the hexagon AEFBGH (VI. 24. cor.) is to the hexagon CIKDLM in the duplicate ratio of the diagonal AB to the corresponding diagonal CD, or (V. 24.) as the square of AB to the square of CD. Wherefore the successive polygons which arise from a repeated bisection of the intermediate arcs, and which approach continually to the areas of their containing circles, must have still that same ratio. Consequently the limiting space, or the circle : AEFBGH, is to the circle CIKDLM, as the square of AB to the square of CD. Cor. 1. It hence follows, that if semicircles be described,


on the sides AB, BC of a right-angled triangle, and on the hypotenuse AC another semicircle be described, passing (III. 19.) through the vertex B, the crescents AFBD and BGCE are together equivalent to the triangle ABC. For, by the Proposition, the square of AC is to the square of AB, as the circle on AC to the circle on AB, or (V. 3.) as the semicircle ADBEC to the semicircle AFB; and, for the same reason, the square of AC is to the square of BC, as the semicircle ADBEC to the semicircle BGC. Whence (V.8. and 19.) the square of AC. is to the squares of AB and BC, as the semicircle ADBEC to the semicircles AFB and BGC. But a (II. 10.) the square of AC is equivalent to the squares of AB and BC, and therefore (V. 4.) the semicircle ADBEC is equivalent to the two semicircles AFB and BGC ; take away the common segments ADB and BEC, and there remains the triangle ABC equivalent to the two crescents or lunes AFBD and BGCE. Cor. 2. Hence the method of dividing a circle into equal portions, by means of concentric circles. Let it be required, for instance, to tri- - sect the circle of which AB is a diameter. Divide the radius AC into three equal parts, from the points of section draw perpendiculars DF, EG meeting the circumference of a semicircle described on AC, join CF, CG, and from C as a centre, with the distances CF, CG, de

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scribe the circles FHI, GKL: The circle on AB will be divided into three equal portions, by those interior circles. Forjoin AF and AG: Because AFC, being in a semicircle, is a right angle (III. 19.), AC is to CD (VI. 15. cor. 1. and V. 24), as the square of AC to the square of CF, that is, as the circle on AB to the circle FHI; but CD is the third-part of AC; wherefore (V. 5.) the circle FHI is the third part of the circle on AB. In like manner, it is proved, that the circle GKL is two third-parts of the circle on . AB. Consequently, the intervening annular spaces, and the circle FHI, are all equal.


The area of any triangle is a mean proportional

between the rectangle under the semiperimeter and its excess above the base, and the rectangle under the separate excesses of that semiperimeter above the two remaining sides.

The area of the triangle ABC is a mean proportional between the rectangle under half the sum of all the sides and its excess above AC, and the rectangle under the excess of that semiperimeter above AB and its excess above BC.

For produce the sides BA and BC, draw the straight . lines BE, AD, and AE bisecting the angles CBA, BAC, and CAI, join CD and CE, and let fall the perpendiculars DF, DG, and DH within the triangle, and the perpendiculars EI, EK, and EL without it.

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The triangles ADF and ADG, having the angle DAF equal to DAG, the angles F and G right angles, and the common side AD,-are (I. 20.) equal; for the same reason, the triangles BDG and BDH are equal. In like

manner, it is proved, that the triangles AEI and AEK

are equal, and the triangles BEI and BEL. Whence the triangles CDH and CDF, having the side DH equal to DF, the side DC common, and the right angle CHD equal to CFD,-are (I. 21.) equal; and, for the same reason, the triangles CEK and CEL are equal. The perimeter of the triangle ABC : B

is therefore equal to twice f
the segments AF, FC, and
BG ; consequently BG is
the excess of the semiperi-
meter above the base AC,
and AG is the excess of
that semiperimeter—or of
the segments BH, HC,
and AG,-above the side
BC. But the sides AB

and BC, with the segments AK and CK, or AI and CL,

also form the perimeter; whence, BI being equal to BL, the part AI is the excess of the semiperimeter above the side AB. Now, because DG and EI, being perpendicular to BI, are parallel, BG: DG. : BI: EI (VI. 2.), and consequently (V. 25. cor. 2.) BIx BG: BIx-DG ::DG x BI: DG x EI. But since AD and AE bisect the angle BAC and its adjacent angle CAI, the angles GAD and EAI are together - equal to a right angle, and equal, therefore, to IEA and EAI ; whence the angle GAD is equal to IEA, and the


right-angled triangles DGA and AIE are similar. Wherefore (VI. 11.) DG : AG :: AI: EI, and (V. 6.) DG x EI = AG x AI ; consequently BIx BG : DG x BI : : DG x BI: AG x AI. But the triangle ABC is composed of three triangles ADB, BDC, and CDA, which have the same altitude; and therefore its area is equal to the rectangle under DG and half their bases AB, BC, and AC, or the semiperimeter BI. Whence the area of the triangle ABC is a mean proportional between the rectangle under BI and its excess above AC, and the rectangle under its excess above BC and that above AB. Cor. Hence the area of a triangle will be expressed numerically, by the square root of the continued product of the semiperimeter into its excesses above the three sides.

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To convert a given regular polygon into another, which shall have the same perimeter, but double the number of sides.

It is evident that, by lines radiating from the centre of the inscribed or circumscribing circle, a regular polygon may be divided into as many equal and isosceles triangles as it has sides. Let AOB be such a sector of the given polygon; from the centre O let fall the perpendicular OC, and produce it to D, till OD be equal to OA or OB, and join AD and BD. The isosceles triangle ADB is therefore (IV. 1.) constructed on the same base with AOB, and has only half the vertical angle. Consequently twice as many of such angles could be constituted about D, as were

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