=

Given two triangles ACB, ADB, having the same base AB, and AC + CB = AD + DB; to prove that if AC CB, the triangle ACB ADB.

(Const.) For draw BE perpendicular to AB,

and draw CF,

KB,

DK perpendicular to BE, make FE FB and KG =

join CE and DG.

(Dem.) Then the triangles CFE,

CFB have BF: FE and FC common,

and the angles at F right angles,

therefore CE = CB.

For a similar

Therefore AC +

and AD + DG =

AD + DB. Now angle ECF = BCF

= CBA (I. 29) = CAB (I. 5),

fore ECF+ ACF:

there

D

CAB+ ACF =

therefore

two right angles (I. 29),

AC and CE are in the same straight line. But AD + DG AG, if

AG were joined, therefore AG ▲ AE, and hence the point G is nearer to the perpendicular AB than E is (1, Cor.), hence GBEB, Therefore the altitude of the triangle ADB being less than that of ACB,

and therefore KB FB.

ACB (VI. 1, Cor. 2).

PROPOSITION III.

THEOREM.

ADB

4

Of all isoperimetrical polygons of the same number of sides, the equilateral polygon is a maximum.

(Const.) Let ABCDE be the maximum polygon, then if two of its sides as ED, DC be not equal, let EF, FC be drawn equal. (Dem.) Then the triangle EFCEDC (2); and therefore the given polygon is less than ABCFE, which is contrary to hypothesis. Therefore ED: = DC;

and it may be similarly proved that any two adjacent sides are equal.

PROPOSITION IV. THEOREM.

Of all triangles having only two sides given, that is the greatest in which these sides are perpen

dicular.

Given two triangles ABC and ABD having the common base AB and AC = AD; to prove that if AC be perpendicular to AB, the triangle ABC¬ABD.

(Const.) For draw DE perpendicular to AB. (Dem.) Then (1) DEAD,

D

A E

and therefore DE

< AC; Cor. 2).

hence the triangle ACD ADB (VI. 1,

Of all the angles subtended at the centres of different circles by equal chords, that in the least circle.

is the greatest.

(Dem.) Since the same chord AB subtends the angles D and C at the centres of two circles,

are the radii, (I. 21).

of which CB and DB

then the angle DC

PROPOSITION VI.

THEOREM.

There is only one way of forming a polygon, all of whose sides are given, except one, and inscribed in a semicircle, of which the unknown side is a diameter.

E

Let the sides of the polygon ABCDEF be all given except AB, and inscribed in a semicircle AEB, of which AB is the diameter. If now a greater circle were taken, the angles at the centre subtended by the chords BC, CD, DE, EF, and FA would be respectively less (Prop. 5) than the angles subtended by them at G, the centre of AEB; that is, they would be less than two right angles;

and therefore the extremities

G

A, B of the given sides would not fall at the extremities of a diameter. If a less circle were taken, the sum of the angles would exceed two right angles, and the same consequence would follow. Hence the polygon in question can be inscribed in only one semicircle.

Schol. The order of the sides AF, FE, &c., may be altered in any manner, the diameter AB and the area of the polygon remaining the same. For the segments cut off by these sides are always the same, and the area of the polygon is equal to that of the semicircle diminished by these segments.

Among all polygons whose sides are all given but one, that is, the maximum whose sides can be inscribed in a semicircle, of which the unknown side is the diameter.

Let ABCDE be the greatest polygon, which can be formed

with the given sides AE, ED, DC, and CB, and the last AB assumed of any magnitude. (Const.) Join AD and DB. (Dem.) Then if the angle ADB were not a right angle, by making it so the triangle ADB would be increased (4), and the parts AED, DCB remaining the same, the whole polygon would be increased. But the given polygon, being a maximum, cannot be augmented;

A

B

therefore the angle ADB must be a right angle. In the same manner the angles AEB, ACB must be right angles, and the polygon is inscribed in a semicircle.

Of all polygons formed with given sides, that which can be inscribed in a circle is a maximum,

Given two polygons, ABCDE and abcde, whose corresponding sides are equal; namely, AB = ab, BC = bc, &c.,

the former be inscribed in a circle, but the latter incapable of being so inscribed; to prove that

the former is the

greater.

(Const.) Draw from A

the diameter AF;

join CF and FD, and

and let

make the triangle cfd CFD in every respect, so that cf = CF,

df = DF, and join af.

(Dem.) The polygon ABCF

in For a similar

abcf (7), unless the latter can be inscribed in a semicircle having af for its diameter, which case the two polygons would be equal. reason the polygon AEDF aedf, unless in the case of a similar exception, by which the polygons would be equal. Hence the whole polygon ABCFDE abcfde, unless the latter can be inscribed in a circle. But it cannot; therefore the former is the greater. Take from both the equal triangles CFD, cfd, and there remains the polygon ABCDE abcde.

Schol.-It may be shewn, as in the sixth proposition, that there is only one circle in which the polygon can be inscribed, and therefore only one maximum polygon; and this polygon will have the same area in whatever order the sides be arranged.

Of all isoperimetrical polygons, having the same number of sides, the regular polygon is the greatest.

and

it is therefore a regular

(Dem.) For (3) the maximum polygon is equilateral, (8) it can be inscribed in a circle,

polygon.

PROPOSITION X.

THEOREM.

Of two isoperimetrical regular polygons, the one having the greater number of sides is the greater.

Let DE and AB be half-sides of these polygons, CB their apothems, lying in the same straight line, and DOE the half-angles

(Const.)

at their centres.
Since these angles are un-
equal, the sides OD, CA,
if produced, will meet in
some point F. Draw
FG perpendicular to CG,

and from the centre
O describe the arcs GI and
MN, with the radius OG
and the radius OM = CG;
centre C with the radius CG.

OE and and ACB

(Dem.) The polygons being regular, their perimeters are the same multiples of their half-sides that four right angles is of their half-angles at their centres, but their perimeters are equal; therefore by direct equality, angle 0:C DE: AB. But O:C MN : GH (VI. 33), therefore DE: AB =

=

=

MN : GH; but OG: OM or CG GI: MN (Qu.* VIII. Cor. 1), therefore (VI. 23, Cor. 1) DE OG: ABCG = MNGI: MN GH = GI: GH (VI. 1) if MN be the altitudes of these two rectangles. But the triangles OED and OGF are similar, therefore OE: OG DE: FG (VI. 4), therefore OE FG: DE OG (VI. 16). Also, from the similar triangles ABC and FGC, CB: AB = CG: FG; and hence CB FG AB⚫CG. Therefore OE FG: CB · FG = GI: GH, or (VI. 1) OE: CB GI: GH. But (Qu. II. Cor. 2) GK GH, therefore GI is still GH, therefore OE CB. But the perimeters of the polygons being equal, their areas will be proportional to their apothems; therefore the polygon, of which OE is the apothem, is the greater. But its half-angle O is less than that of the

* Qu. refers to the preceding book on the Quadrature of the Circle.

other, or it has a greater number of sides;

therefore the

polygon having the greater number of sides is the greater.

PROPOSITION XI.

THEOREM.

T

The circle is greater than any polygon of the same perimeter.

Let AG be a half-side, and C the centre of a regular polygon of the same perimeter as the circle, of which F is the centre, and DE an arc subtending an angle DFE equal to the angle ACB subtended by the side AB;

:AG GC: DHHF GC: HF.

GC: HF AB

IHK be a side of the corresponding circumscribed polygon, I then by the similar triangles ABC and IFK, and therefore P: C DE: IK.

=

or DE: IK; But (Qu. IL Cor. 2) DH IH, and therefore IK DE, hence CP. Now P is a regular polygon, and of all isoperimetrical polygons having the same number of sides, the regular polygon is the greatest; but the circle has been proved to be greater than this polygon; it is therefore greater than any isoperimetrical polygon of equal perimeter.

PROPOSITION XII. THEOREM.

Of all polygons having the same area and the same number of sides, the regular polygon has its perimeter a minimum.

Let A and P be the area and perimeter of a regular polygon B, and N the number of its sides, and A', P', N' those of an irregular polygon B', if A A' and N N', then PZP'.

P" and N" If P" P and parts of B" be

For let B" be a polygon similar to B' and A", its area, perimeter, and the number of its sides. N" N, then A" A (9); and therefore if the proportionally increased till A"= A, then its perimeter (VI. 20). But P'P", and therefore P'⇒ P.

= P