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In the construction of the figure it is necessary to show, 1. that / CGB = _ ADB, 2. that < CGB = L CBG, 3. that :. BC = CG, 4. that cb, BI, IG, GC = each other, 1 and .. ci is a 5. that om ci is rectangular, (square, i. e., CB 6. that similarly uy is a square, i. e., Ac.

The Demonstration itself consists in showing that AB’, i.e., fig. AE = AC + CBS + figs. Ag + ge, and that these are = aC + CBS + 2 AC X CB.

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PROPOSITION V. Theorem. If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle 4 contained by the unequal parts,

M together with the square of the line between the points of section, is equal to the square of half the line.

Proved by showing that bc = figs. CM + HG +LF; and that these = figs. AH + LF; and these = AD X DB + cd.

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part of it produced, together with the square of half the line bisected, is equal to the square of the line made

up

of the half and the part produced.

Proved by showing that cd' = figs. CH + BF + LG; and these – AL + CM + LG, which = AM + LG, and that these last = AD X DB + cb?.

PROPOSITION VII.

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Theorem. If a straight line be di. vided 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 line and that part, together with the square of the

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other part.

Proved by showing that ABS + BC* = figs. AG + GE + 2 ch + KF, and that these 2 AH + KF; and that these = 2 AB X BC + AC%.

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PROPOSITION VIII. Theorem. If a straight 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 line made up of the whole and

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that part.

In the construction of the figure it is necessary to show that BD = GK, in order to prove that the square BN = square GR, and also to show that BK = KR, in order to prove that Om AK = OM MR.

The Demonstration itself consists in showing that AB + BC", i. e., the fig. AF = figs. AK + MP + PI + KF + BN + GR + xh, and that these are = 2 AK + 2 MP + 2 GR + xh; and that these = 4 AK + XH, and that these last = 4 AB X BC + Aco.

PROPOSITION IX.

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Theorem. If a straight line be divided into two equal and also into two unequal parts,

the
squares

of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.

In the construction of the figure it is necessary to prove,1. that _ AEB is a right Z, which is done by show

ing that each of the ZS CEB, AEC = į rt. Z, 2. that ge = yg, by showing that ZS GEF, EFG,

each = į rt. Z, and :: = ea. other. 3. that DF = DB, by showing that , B and BFD

each = } rt. Z, and :: = ea. other. The Demonstration itself consists in showing that AD' + DB' = AF?; and that this = ACS + CES + EGR + GF, and that these are = 2 c + 2 cdo.

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PROPOSITION X. Theorem. If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line, and of the square of the line made up of the half and the part produced.

In the construction of the figure it is necessary to show,1. that ZS BEF + EFD < 2 rt. Zs, in order to prove

that EB, FD will meet, if produced, as in G; also 2. that _ AEB is a rt. Z,

by showing that each of the

ZS CEA, CEB = { right L, 3. that BD = DG,

s by showing that ĻS DBG, DGB each

- š right 2, and :: = each other. 4. that GF = FE, by showing that _ GEF = _ FGE.

The Demonstration itself consists in showing that ADS + DB% = AG?; and that this = AC? + ce + EF + FG', and that these = 2 Aco + 2 cd'.

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PROPOSITION XI.

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Problem. To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square of the other part.

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Steps of the Demonstration. 1. Prove that cF X FA + AE – EF', and that this 2.

= AE + AB, 3. that cr X FA, i.e., fig FK = AB', i. e., fig. AD, 4.

that FH = HD, 3.

that . AB X BH = AH'.

PROPOSITION XII. Theorem. In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle, between the perpendicular and the obtuse angle.

Obs. This proposition is so short, and each step so necessary to the proof, that to attempt to simplify it would be useless. The learner will find no difficulty in it whatever, if he can remember that bco + CD + 2 BC XCD is to be substituted for DB; but this should be carefully noticed, as it is very likely to escape the memory.

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PROPOSITION XIII. Theorem. In every triangle, the square of the side subtending any of the acute angles is less than the

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