Scholium. The algebraic formulas,

(p + q)2 = p2 + q2+2pq, ( p − q)2 = p2+q2 — 2pq,

which have been made the bases of Theorems XV. and XVI., and the other well-known formula,

being translated into geometrical language, will give rise to some new Theorems in regard to areas, with which we will close this Book.

THEOREM XXX.

The square constructed on the sum or on the difference of two lines, is equivalent to the sum of the squares constructed respectively on these lines, plus or minus twice their rectangle.

The simple inspection of the figure is nearly sufficient to satisfy one of the truth of this double proposition.

First. Let AE be the greater of the two lines, EB the less. Construct the squares AEIG, ABCD, and produce EI until it meets CD at F.

L

M

D

F

K

G

I

E

B

The square constructed on AB, the sum of the two lines, is evidently composed of the squares AEIG, IKCF, constructed on the line AE and on the line IK = EB, increased by the two rectangles BEIK, GIFD, which have for their respective bases GI = AE, EI = AE, and for altitudes GD = EB, IK = EB. Hence, we have

square AB = square AE + square EB + twice rectangle AE×EB.

Secondly. Let AB be the greater line, BE the less, that which gives AE for the difference of these two lines. Construct the same figure as in the first case, and, in addition, construct the square GDLM equal to IKCF.

The square AEIG is equal to the square ABCD plus the square GDLM, minus the two rectangles EBCF, MIFL. Now, these two rectangles have respectively for bases FE = AB, IM = GK = AB, and for altitudes BE IKIF.

Hence we have

=

square AE= square AB+ square EB- twice rectangle ABxBE.

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THEOREM XXXI.

The rectangle constructed on the sum and the difference of two lines, is equivalent to the difference of the squares constructed on these two lines.

Let AB be the greater line, BEBE' the less, so that AE will represent the sum of these two lines, and AE' their difference. Construct on AE as a base, and AGAE' as an altitude, the rectangle AENG, also the square ABCD, and draw E'IF perpendicular to AB, forming the square AE'IG.

G

D

F с

N

I

K

E' B

E

The two rectangles BENK, GIFD are equal, having equal bases and equal altitudes, namely, BK = AG=GI= AE', EB =E'B= IK = IF=DG; hence it follows that the rectangle AENG is equivalent to the figure DFIKBA. But this figure is the difference of the squares constructed on AB and on IK = BE' = BE.

Hence, we have

=

rectangle AENG square AB - square BE.

FOURTH BOOK.

THE PROPORTIONS OF LINES AND THE AREAS OF FIGURES IN CONNECTION WITH THE CIRCLE.

DEFINITIONS.

I. Similar arcs, Sectors, or Segments, are those which, in different circles, correspond to equal angles at the centre.

II. When two similar sectors are superposed so that their equal angles coincide, their difference is called a circular trapezoid. Thus the space BB'C'C is a circular trapezoid.

B

B'

III. The space included between two concentric circumferences is called a circular

ring.

O

IV. The arc of a circle is said to be rectified, when it is devel oped; that is, unfolded or drawn out into a straight line. It is only in this rectified form that we can conceive of its length, since it could not otherwise be compared with the linear unit.

PROPORTIONAL LINES CONNECTED WITH THE CIRCLE.

THEOREM I.

When two chords intersect each other within a circle, the segments will be reciprocally or inversely proportional.

Let the two chords AA', BB' intersect at P. Drawing the auxiliary chords AB', A'B, we thus obtain two triangles PAB', PA'B, which are similar, since the angles A at P are equal (B. I., T. I.); the angles at A and B are equal, so also are the angles.

at A' and B' (B. II., T. X.), hence these

B

E

triangles are equiangular, and consequently similar (B. III., T. VI.), and their homologous sides give this proportion,

PA: PB:: PB' : PA'.

Scholium I. When one of the chords

is a diameter, and the other is perpendic

ular to it, we have

PA: PB:: PB:PA'.

The line PB, drawn perpendicular to a diameter, is called an ordinate to this diameter. Hence, we have this condition:

A

E

P

B

B'

In a circle any ordinate to a diameter is a mean proportional between the two segments of this diameter.

We can show that this property is one of the consequences of the properties of right-angled triangles; because if we join A, A' with the point B, we shall form a right-angled triangle (B. II., T. X., C. I.), which gives (B. III., T. XIII.) this proportion,

PA: PB:: PB: PA'.

Scholium II. This same right-angled triangle ABA' gives the proportion,

That is,

AA': AB::AB: AP.

Any chord which is drawn through the extremity of a di-. ameter is a mean proportional between its projection (B. III., T. XIV., S. II.) on the diameter and the diameter itself.

THEOREM II.

When two secants intersect each other, without a circle, they will be reciprocally proportional to their external segments.

Let the two secants intersect each other at P. Draw, as in the preceding Theorem, the chords AB', A'B, and the two triangles AB'P, BA'P, will be similar, since the angle at P is common, and A =B, each being measured by half of the arc A'B' (B. II., T. X.), and the homologous sides give the proportion

PA: PB:: PB': PA'.

A

P

A'

B'

E

B

THEOREM III.

When a secant and tangent are drawn from the same point, the tangent is a mean proportional between the secant and its external segment.

This Theorem is in reality only a particular case of the preceding, when the points B, B' of the secant PB, are united in one point.

But we will give a direct demonstration, by drawing the chords AB, A'B. We have, in effect, two triangles, PAB, PA'B, similar, since the angle at P is common and the two angles at A and B are equal (B. II., T. X., C. II.).

Hence, we have this proportion,

P

E

B

PA: PB:: PB: PA'.