rake APPENDIX 
TO THE ELEMENTS OF GEOMETRY. 
SECT, I. 
Or THE Maxnta AND Minima or GrometTnicat 
QuanTITIES. 
Definitions. 
1. A quantity is said to be a maximum, when it is the 
greatest ; and a minimum, when it is the least of all 
‘nL. quantities of the same kind. 
Thus the diameter of a circle is a mazimum among all 
the chords that ean be drawn in a circle ; and the per- 
ndicular is a minimum among all lines that can be 
rawn from a given point to terminate in a straight 
line. ; 
2. Figures are called tsoperimetrical, when they have 
equal perimeters. ? 
Prop. I. Tueor. 
Of all triangles having the same base and the same 
perimeter, the maximum triangle is that in which the 
two indeterminate sides are equal. 
Let ACB, AMB, be two triangles on the same base 
AB, and such that AC-+-CB=AM-+MB;; thenif AC= 
CB, and AM, be greater or less than MB, the triangle 
ACB is greater than AMB. 
Draw CE perpendicular to AB, and MD perpendieu- 
lar to CE; join AD, BD, and in AD produced take 
DF=DB, and join MF, Then AE=BE (Geometry, 
Schol. to 12 of I. Part I.) and AD=BD (6. 1.) and 
the angle FOM=DAE (3 Cor. 21. idea (2. 1)= 
BDM ; hence the triangles FDM,BDM are equal, (6. 1.) 
-and MF=MB, and AM-+- MF=AM+.MB=AC4.CB= 
2AC; but AM4MF is greater than AF, or 2AD; 
therefore 2 AC>2AD, and ACAD; and hence 
EC—ED (16. 1.) Now EC and ED are the altitudes 
of the triangles ACB and AMB respectively ; therefore 
(C es) ACB is greater than the triangle AMB, 
or. 6, 4). . 
Prop, II. THEoR. 
- Of all isoperimetrical poly, of the same number 
of sides, that which is a maximum has its sides equal. 
For let ABCDEF be the maximum polygon, if the side 
BC be not equal to CD on the base BD, make an isos« 
celes triangle BOD, which shall be isoperimetrical to 
BCD ; then the triangle BOD is greater than BCD (1.), 
and consequently the poly, ABODEF is 
the polygon ABCDEF ; therefore this last is not the 
greatest of all polygons having fhe same number of sides 
and the same er, which is contrary to what we 
have sup 3 hence BC must be equal to CD; and 
in like manner it may be demonstrated, that any two 
adjoining sides are equal. 
Prop. III. Tuzor. 
Of all triangles constructed with two given sides 
which contain any angle, that is the greatest, of which 
the given oe contain a right angle, aah 
GEOMETRY. 
235 
Let BAC, BAD, be two: triangles, which have the 
side AB common, and AC=AD. If BAC be a right 
angle, the triangle BAC shall be greater than BAD, or 
BAD’: for the triangles BAC, BAD, are to one ano- 
ther as their altitudes AC, DE, because they have the 
same base ; but DE is»less than AD, or its equal AC ; 
therefore the triangle BAD is less than BAC. 
Prop. IV. Turor. 
~ If all the sides of a polygon be given, except one, 
the poly ‘on will be a maximum when all its angles are 
on half the circumference of a circle of which the un- 
known side is the diameter. 
Appendix- 
——— 
Fig. 172. 
Let ABCDEF be the greatest polygon that can be Fig, 17%. 
formed by the given lines AB, BC, CD, DE, EF, and 
the indetermined line AF. Draw AD, FD to the ver- 
tex of any one of its angles. If the angle ADF be not a 
right angle, supposing two parts of the polygon, ABCDA 
and FEDP, to remain the same, the triangle ADF, and 
consequently the whole polygon, might be increased by 
making ADF a right angle (3.) ; but the polygon be- 
ing by hypothesis a maximum, it cannot be increased, 
therefore the angle ADF must be already a right angle. 
The same is also true of the angles ACF, ABE, AEF ; 
therefore all the angles of the polygon are in the cir- 
cumference of a semicircle of which AF is the diame- 
ter, ‘ 
Scnotrum. This proposition gives rise to a question, 
whether it be possible to form different polygons which 
. shall each be inscribed in a semicircle, and have all their 
sides, except that which is the diameter, equal to given 
lines? Before deciding this question it may be obser- 
ved, that if one and the same chord AB (Fig. 174.) 
subtends ares described with different radii AC, AD, 
the angle which the chord subtends at the. centre of 
the greater circle shall be less than the angle it sub- 
tends at the centre of the less circle: For the angle 
ADO = ACD + DAC; therefore ADO>ACO, and 
doubling each ADB=ACB. . 
Prop, V. Tu20n, 
There is only one way of forming a 
ABCDEF which shall be inscribed in a semicircle, an 
have all its sides, except the diameter AF, equal to 
given lines. 
. Fer supposing a circle to be found that satisfies 
the question, if a greater circle would also satisfy it, 
the chords AB, BC, CD, &e. would subtend lesser 
angles at the centre of this than at the centre of the 
other circle; and the sum of these angles would be 
less in the one than in the other circle; but by the 
nature of the figure, in each the sum should be the 
same, viz. two right angles, therefore the polygon can« 
not be inscribed in two different semicircles. 
Scuonrum. The order of the given sides AB, BC, 
CD, &c. may be changed, and still the diameter of the: 
circle shall be the same, as well as the area of the poly- 
gon ; for whatever be the order of the arcs AB, BC, 
CD, &c. it is sufficient that their sum be a semicircum- 
ference. The different polygons will also have equal 
areas, because by drawing lines to the centre, the tri- 
angles which’constitute any one polygon will be respec- 
py equal to those which constitute any other, as is 
evident, 
Fig. 174. 
polygon Fig. 173. 
