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(In fig. 1, the ms are equal, because each is double of ▲ BDC.)

Steps of the Demonstration, (figs. 2. and 3.)

1. Prove that AD = EF,

that whole or remainder AE = whole or remainder DF *,

2.

3. 4.

that

that A EABA FDC,

Ac1 EC.

PROPOSITION XXXVI.

Α

Theorem. Parallelograms upon A equal bases, and between the

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E

B

PROPOSITION XXXVII.

Theorem. Triangles upon the same bases, and between the same parallels, are equal.

*The words" whole or remainder" are used in order to suit both figs. (2 and 3)-since, in fig. 2, DE is added to AD and EF, and in fig. 3, is taken away from AD and EF.

Steps of the Demonstration.

1. Prove that Om FCTM FB,

2.

that the As are respectively = these equal □ms, and :: = each other.

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m

TM HE,

that the As are respectively = these equal ms and each other.

PROPOSITION XXXIX.

(Argument ad absurdum.)

Theorem. Equal triangles upon the same base, and upon the same side of it, are between the same parallels.

Steps of the Demonstration.

Suppose that AD is not || BC,

but that some other line, as AE, is || BC.

Then, on that supposition, prove,

1. that ▲ ABC = ▲ EBC,

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PROPOSITION XL.

(Argument ad absurdum.)

Theorem. Equal triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels. B

D

C E

Steps of the Demonstration.

Suppose that AD is not || BF,

but that some other line, as AG, is || BF, Then prove, on that supposition,

1. that ▲ ABC = A GEF,

2. that ▲ GEFA DEF, i. e., less

3. that. AD || BF.

B

greater, which

shows the supposition to be false,

PROPOSITION XLI.

E Theorem. If a parallelogram and a triangle be upon the same base, and between the same parallels; the parallelogram shall be double of the triangle.

Proved by showing that

2 A EBC.

BD = 2 ▲ ABC, which

PROPOSITION XLII.

Problem. To describe a parallelogram, that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

A F

G

b

B

E

Steps of the Demonstration.

1. Prove that whole ▲ ABC = 2 ▲ ACE,

2.

that FCA ABC (and it has CEF= ZD by construction.)

PROPOSITION XLIII.

Theorem. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.

E

A H

D

K

F

B G

Steps of the Demonstration.

1. Prove that ▲ ABC

2.

3.

A ACD,

that A AEK + ▲ KGC = ▲ AKH + ▲ KCF, that complement вK complement KD.

PROPOSITION XLIV.

Problem. To a given straight line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

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In the construction of the figure, it is necessary to prove that S BHF + HFE < 2 rights, and that consequently HB and FE will meet; and to state that FL is a of which LB, BF are complements of about the diameter HK.

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ms

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Problem. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.

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In the construction of the figure it is necessary to show,

1. that FKH = GHM,

2. thats FKH +KHG ≈ ≤S KHG + GHM,

3. that S KHG + GHM=2 rights,

4. that кH, Hм are in one right line.

in order

to

prove,

(In a similar manner it must be proved that FG,

GL are in one right line.)

Also that FM is a m.

The Demonstration itself is merely the showing from the construction that the whole

whole

rectilineal figure, because the parts of the one = the parts of the other.

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