Let ACBD be the inscribed quadrilateral, and we shall have

AB × CD = AC × BD + AD × BC.

For, drawing CE, making the angle ACE equal to the angle BCD, and consequently the angle ECB equal to the angle ACD.

The two triangles CEA, CBD are similar, since the angle ACE equals the angle DCB, by construction, and CAE = CDB, since each is in the same segment (B. II., T. X.). Their homologous sides give this proportion,

consequently,

AC: CD::AE:BD;

CD × AE = AC × BD.

The comparison of the homologous sides of the two triangles BEC, DAC, which are similar, gives

and

AD: BE:: CD: BC,

CD × BE = AD × BC.
X

Adding the corresponding members of these equations we obtain

or

CD × (AE+BE) = AC × BD +AD × BC,
AB × CD = AC × BD + AD × BC.

Scholium. This Theorem has many important applications. FIRST. To find the chord AB of the sum of two arcs AC, CB, when their chords are known.

For abridgment, let a, b, c denote the three sides BC, AC, AB of the triangle ABC, and R the radius of its circumscribing circle. Draw the diameter COC'= 2R, also the supplementary chords AC', BC'.

Now the inscribed quadrilateral ACBC' will give

AB × CC'=AC × BC'+AC'× CB.

But the triangles CAC', CBC' being right-angled, since they are each in a semicircle, give

AC′= √CC22 — AC2, BC'✓CC22 — BC2 ;
- ;

and since CC'=2R, AC=6, BC = a, the foregoing expressions will become

AC'= √4R2 — b2, BC'=√4R2 — a2,

these values, substituted in the first condition, cause it to become

c × 2R = b√4R2 — a2 + a √4R2 — b2

SECONDLY. To find the chord AB = c, of double an arc, when we know the chord BC = a of the arc itself.

This will evidently be accomplished by making b = a in the general expression just formed; so that, in this case, we have

C =

α R

√4R2 — a2.

(2)

C'

THIRDLY. To find the chord of half an arc, when we know the chord of the arc itself.

or,

In the triangle CBC', CB2== CI × CC' (B. III., T. XIII.).

Now, CI=OC-OI = OC — √OB2 — BI2 (B. III., T. XIV.) ;

CI = R— √R2 — 1c2.

Hence, CB2=a2=2R×(R-√R2-4c2)=2R2-R√4R2—c2;

It would not be difficult to deduce, algebraically, the value of a in terms of c and b, from formula (1); that is, to find the chord of the difference of two arcs, when we know their chords. When found, the expression is

We would also remark that formula (3) could have been obtained very readily from (2) by the ordinary rules of algebra.

DETERMINATION OF THE SIDES AND OF THE AREAS OF REGULAR POLYGONS.

THEOREM V.

The area of any regular polygon is equal to half the product

of its perimeter by its apothem.

For, the equal isosceles triangles OAB,

B

area ABCDEF=(AB+BC+ CD + etc.) × 1OG; or, more concisely,

A = {p × r.

A denoting the area of the polygon, p its perimeter, and r the apothem, or the radius of the inscribed circle.

Scholium. Calling R the radius of the polygon, n the number of sides, and a the length of one of the sides, we have, from the right-angled triangle OGA,

2

r=

OG = √OA2 — AG2; that is, r = √R2 — 4a2.

The perimeters of two regular polygons of the same number of sides are proportional to the radii of their inscribed circles, or of their circumscribed circles; and their areas are proportional to the squares of those radii.

Since these polygons are regular, and have the same number of sides, they are similar figures. Consequently, the isosceles triangles which have their vertices at the centre, and for bases

the sides of the polygons, are each to each equiangular and similar. The radii of the inscribed circles and of the circumscribed circles are homologous lines. Therefore the proposition is true (B. III., T. XXVIII., C.).

Let R and R', r and r', a and a', p and p', A and A', denote, respectively, the radii, the apothems, the sides, the perimeters, and the areas of these two polygons, and we shall have as follows:

pp': RR ::rr' ::a: a',

In a given circle, knowing the side of a regular inscribed polygon, we can obtain :

I. The value of the side of a regular inscribed polygon of half the number of sides.

II. The value of the side of a regular inscribed polygon of double the number of sides.

III. The radius and the side of a circumscribed polygon similar to the given polygon.

First. Since AC subtends an arc double the arc subtended by AB= a, it is sufficient to substitute AC for c in formula (2), scholium to T. IV. Which thus becomes

Secondly. We obtain, in the same way, the value of AI, by means of formula (3), by substituting AI for a, and a for e. Making these substitutions we have

AIV2R2-R √4R2 — a2.

(2)

Thirdly. As to the values of OM and MN, the two similar

triangles OMN, OAB, of which OI and OK are homologous

but we have OA=OIR, AB = a, and

OK = √OA2 — AK2 = √R2 — 1α2 — ≥√4R2 — a2.

=

Scholium. Knowing the radius and the side of each of the three new polygons, we are able to deduce the values of their areas by substituting, in the formula (T. V., S.),

for n, a, R, the respective values just obtained.

used for n, for the first polygon, and 2n for n, for the second. It remains the same for the third.

THEOREM VIII.

The area of a regular inscribed polygon, and that of a regular circumscribed one of the same number of sides being known, we are always able to obtain the areas of two regular polygons, the one inscribed and the other circumscribed, of twice the number of sides.

Let A and B denote respectively the areas of the inscribed and circumscribed polygons, each having a number of sides denoted by n, and A' and B', the areas respectively of the inscribed and circumscribed polygons of double the number of sides.

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I m

n

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B

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