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

D

B

(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,

E D

HKB

CG, and from C as a centre, with the distances CF, CG, de

3

scribe the circles FHI, GKL: The circle on AB will be divided into three equal portions, by those interior circles. For join 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.

PROP. XXIX. THEOR.

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.

B

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 is therefore equal to twice the segments AF, FC, and BG; consequently BG is the excess of the semiperimeter 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

I

H

G

A

K

F

E

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.) BI× BG: BI× DG :: DG × BI : DG × 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 × EI = AG X AI; consequently BI× BG: DG × BI :: DGX BI: AGX 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.

PROP. XXX. PROB.

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

AD and BD in E and F, and the straight line joining these points must (VI. 2.) be equal to half the base AB. Wherefore the triangle EDF, repeated about the vertex D, would form a regular polygon with twice as many sides as before, but under the same extent of perimeter, since each of those sides has only half the former length.

Cor. 1. Hence DG, the radius of the circle inscribing the derived polygon, is half of CD, that is, half of the sum of OC and OA, the radii of the circles inscribing and circumscribing the given polygon. Again, since AOD is evidently isosceles, AD2=20A.CD (II. 23. cor.), and consequently DE the radius of the circumscribing derived polygon, being the half of AD, is a mean proportional between OA and DG, the radius of the circle circumscribing the given polygon, and the radius of the circle inscribing the derived polygon.

Cor. 2. Hence the area of a circle is equivalent to the rectangle under its radius, and a straight line equal to half its circumference. For the surface of any regular circumscribing polygon, being composed of triangles such as EDF, which have all the same altitude DG, is equivalent (II. 5.) to the rectangle under DG, and half the sum of their bases, or the semiperimeter of the polygon. Therefore the circle itself, since it forms the ultimate limit of the polygon,