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common difference : if there are three quantities, for instance, and the first exceeds the second by as much as the second exceeds the third, such quantities are in arithmetical progression,—the first and third being named the extremes, and the second the mean. Take, in the st. line AB, A D the greater segment = 16, DB the less 4, and AC the half sum = 10. AC is the mean between AD and DB, the common difference being the intermediate part CD. In numbers we may take 16, 10, 4;

these are in arithmetical progression, because their common difference is 6. Generally we affirm—The arithmetical mean is half the sum of the extremes, and the common difference is half the difference of the extremes."

“The fifth proposition may then be announced thus :The square of the arithmetical mean is equal to the rectangle under the extremes together with the square of the common difference.”—LARDNER's Euclid, p. 71.

3. This proposition is employed to establish one of the most important properties of st. lines cutting one another within a circle ; as Prop. 35, bk. iii.

If two straight lines cut one another within a circle, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other."

4. By this proposition the method may be demonstrated of finding the value of an Adfected Quadratic Equation in Algebra, that is, of an equation which con tains the square and the first power of an unknown quantity. In such equations the square is not completed, being defective by the square of half the co-efficient of the second term. The problem to be solved is—Ĝiven, as in the last fig., in the square CF, LG and 2 CH, required DM ; i.e., required the magnitude which will complete the square. The magnitude DM is the square on DB, and D B is equal to DH ; and 2 DH.CD represents 2 CH. In the magnitude 2 DH.CD, 2 DH is the factor or co-efficient of CD; and if we take half of that co-efficient we have DH,—the square of which, J) M, is the magnitude we are seeking to make up the square. To solve an adfected quadratic equation, as 22 + 4x = 12, we complete the square by adding to each side the square of half the co-efficient 4; then 2 + 4x + 4 = 12 + 4 = 16; extracting the square root,

+ 2 =4..x = 2.

PROP. 6.- THEOR.

If a st. line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together with the square of half the line bisected, is equal to the square of the st. line which is made up of the half and the part produced.

Cons.—46. I., Pst. 1, and 31. I.
DEM.—43. I., Ax. 2, 36, I., Ax. 1, Def. 30. I., Cor. 4. II.

Exp. 1 | Hyp.
Let! A B be bisected

с B D
in C and pro-
duced to D,

L H

M 2 Concl.

then 0 AD. DB K
+

the square
on CB the

E G F Cons. 1 by 46. I. Pst. 1. On st. line CD construct the square CEFD,

and join DE; 2 31. I.

through. B draw | BHG || CE, or DF; 3

.H KLM || AD, or EF; 4

and, A AK || CL, or DM; 5 Concl.

then o AM + LG

square on CD.

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square CF.

=

Dem. 1 by H. & 36. I.::|AC = CB, O AL = CH;

2 | 43. I. & Ax. 1. but o CH = HF;:.0 AL = HF; 3 Add. & Ax. 2. Adding to each oCM,

then AM = the gyomon CMG. 4 Cons.,Cor.4.11. But | DM - | DB; .. AM contained by

AD.DM = 0 AD.DB;
5 Ax. 1.

and .. gnomon CMG = DAD.DB:
6 Add. Cor. 4. II. Adding LG the square on CB to each,
7 | Ax. 1 & 2. then a AD.DB and square on CB

gnomon CMG and square LG:
8 Const. But CMG and LG make up fig. CEFD

the square on CD; 9 Ax. 2.

:.0 AD. DB + C B’ = CD?. 10 Recap.

Wherefore, if a st. line be bisected and pro duced, &c.

Q.E.D. Alg. & Arith. Hyp.— Let AC or CB=a=8; AB=2 a=16 and BD=m=4

AD=2 a+m=16+4 ; and CD=a+m=8+4.
Alg. Now AD=2 a + m (mm)

and therefore (2 a + m) m=2 am + m2. (Add a?)
therefore (2a + m) m + a2 =a+ 2am + m2
But a2 + 2am + m2 = (a + nu)?

therefore (2a + m) m+a? -= (a + m)? Arith. Now AD=16+4 (x 4)

then (16+4) x 4=(2 x 8 x 4) + (4 x 4) =64+16. (Add 8 x 8)
and 4 (16+ 4) + 64=64+64 +16
But 64+64 +16=(8 + 4)2 =144
therefore 4 (16+ 4) +64=(8.44)2 =144

COR.If a st. line A D be drawn from the vertex A of an isosceles triangle to the base or its production, the difference between the squares of this line and the side of the triangle is the rectangle under the L! segments B D x DC of the base.

For (by Cor. 4, P. 47. I.) AD ~ AC = CE* ~ DE'; but (by 5 and 6. II.) CE?~DE BD.DC, .. AD? ~ AC? = BD. DC.

N.B.-If AD coincide with the perpendicular A E, the part DE will vanish.

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Sch.—The alegebraical results of Prop. 5 and 6 differ only in appearance,

arising from the two ways in which the difference between two unequal lincs may be represented yeometrically when they arc in the same direction.” In Prop. 5, the difference DB=u- m=10 - 6. of the two lines AC=a=10 and CD=m=6, is exhibited by producing the less CD, and making CB=CA. Then DB = AC or CB-CD=a m=10 - 6=4. In Prop. 6, the difference D B of the two unequal lines CD and CA, is exhibited by cutting off from CD, the greater, a part C B equal to CA, the less ; then D B =CD-CB or CA=(a + m) - a=(8 + 4) - 8=12 – 8= 4.

USE AND APP.—MAUROLICO, a mathematician,

A and abbot of Messina, who died A.D. 1575, measured the diameter of the earth by making use of this proposition.

F From A the top of a mountain, the height of

В.

D which A D is known, the angle C A B is measured, formed by E A and A B, a tangent to the circle in B, the limit of vision. In the rt. angled triangle ADF, the angles being obtained, the sides A F, FD will be found by Trigonometry; and FV equals

C FB: thus A B=AF+F B, and its square=A Bạ.

Now (by Prop. 6. II.) ED being divided in C and produced to A, the rectangle EA. AD+CD2 =AC?; and (by Prop. 18. III.) A B C is a right angle : therefore AC? = AB? + BC? or CD?, and

E thus the rectangle E A. A D+CD2 = A B’ + C D’; take away the common part CD?, and the rectangle E A.A D=A B’. Divide both sides of the equation by AD, and EA= Then EA-AD=ED,

AD the diameter of the earth.

Example. A mountain is 24 miles, AD, above the level, FD, of the sea ; and the limit of vision, or A B, is 141 miles: required the earth's diameter, DE.

AB?
By the result just obtained, E A= _ and ED=EA-AD.

AD
A B 2
Therefore
141 x 141 19881

=7952.4 miles ; and 7952.4 - 2.5= AD

2.5

2.5 7949.9, the earth's diameter.

A B 2

or

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PROP. 7.-THEOR. If a st. line be divided into any two parts, the squares of the whole line and of one of the parts are equal to twice the rectangle contained by the whole and that part, together with a square of the other part.

CONS.—46. I., Pst. 1 and 31. I.

DEM.—43. I., Axs. 2 and 6, Def. 30. I., Cor. 4. II., Ax. 8, and 34. I. EXP. 1 Нур. Let | A B be divided

A

C B into any two parts

in.C;
2 Concl.
then the squares on H

IK
A B and CB=twice

AB.BC + A C. Cons.1 by 46. I. Pst. 1. On AB make a square

A DE B, and join D

DB; 2 31. 1.

through . C draw CF|BE, 3

and .G HK AB; 4 Concl. then the squares AE + CK = twice AK + HF, Dem. 1 by 43. I. : the compl. AG the compl. GE;

2 Add. & Ax. 1. on adding CK to each, AK = OCE; 3 Ax. 6.

.. AK + OCE - twice O AK;
4 Const. But AK + CE make up gnom. AKF + OCK;
5 Ax. 1. :. AKF and CK together = twice AK:

::
6 D. 30 I., 4. II. but BK = BC,
7 Ax. 6. twice O AK = twice AB. BK,

and twice AB. BK twice AB. BC;
8 Ax. 1 & 8. .. the gnom. AKF + CK twice the

rect. AB. BC; 9 Add. & 34. I. Adding to both the equals HF, i.e., the

square on HG or AC, 10 Ax. 2.

then AKF + CK + HF twice AB. BC

and AC squared:
11 Const.
But AKF + CK + HF make

up

the figures ADEB and CK, 12 C. 1, 3. and A DEB and CK are the squares on AB

and CB, 13 Concl.

:. AB+ BC = twice AB.BC + AC?. 14 Recap.

Wherefore, if a st. line be divided into any two parts, &c.

Q.E.D.

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Alg. & Arith. Hyp.—Let AB = a linear units = 16 ; AC = m = 9; and CB

=n=7. Alg. Then a=m+n; Squaring, a?=m? +2 mn+n?.

(Add na) a2 +n? =m2 +2mn+2na. But 2mn + 2n?=2 (m+n) n=2an,

therefore a? +no=2an+ m2.
Arith. 16=9+7; Squaring, 256=81+126 + 49.

(Add 49) 256 + 49=81 +126 +98.
But 126 +98 = 2 (9+7) 7=2 (16 x 7) = 224,

therefore, 256 + 49=222 +81=305. Another form of stating the same result is,

Let AB=a=16; AC=b=9; and BC=a-b=16 – 9=7.
then A Bo=a?

And 2 AB. BC=2 a? - 2 ab
BC?=a? - 2 ab +62

AC=

63 Sum 2 a? 2 ab +62

2 a2 - 2ab +62

COR. 1. — If AB and BC be considered as two independent lines, AC being their difference, the sum of the squares of any two st. lines is equal to twice the rectangle under them together with the square of the difference ;” i.e., A B+ BC = 2 AB, BC + AC?; or 100+ 64 = (2 x 80) + 4.

CoR. 2.- Hence and (by 4. II.) the square of the sum of two st. lines, the sum of their squares, and the square of their difference, are in arithmetical progressionthe common difference being twice the rectangle under the sum.

By 4. II. (AB + BC)2 = A B + BC + 2 AB. BC; and by Cor. 7. II. AB? + BC = (AB-BC) + 2 AB. BC; or AC + 2 AB. BC. Thus the common difference is 2 AB. BC; therefore the quantities (AB + BC), (ABP + BC?), and AC are in arithmetical progression; as 324, 164, and 4,—the com, dif. being 160.

+

PROP. 8.-THEOR.

If a st. line be divided into any two parts, four times the rectangle contained by the whole line and one of the parts together with the square of the other part, is equal to the square of the st. line which is made up of the whole and that part.

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