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

Steps of the Demonstration, (figs. 2. and 3.) 1. Prove that AD = EF,

that whole or remainder AE = whole or

remainder DF*, that A EAB = A'FDC, that " AC = " Ec.

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PROPOSITION XXXVI. Theorem. Parallelograms upon A equal bases, and between the same parallels, are equal to each other.

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Steps of the Demonstration. 1. Prove that BC = EH,

that Ec is a om,
that om AC = OMEG, (for they are each

= OM EC.)

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 an and EF.

Steps of the Demonstration. 1. Prove that OMFC = OM FB, 2. that the As are respectively = these equal

oms, and :: = each other.

PROPOSITION XXXVIII.

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

Steps of the Demonstration. 1. Prove that OmGc= OM HE, 2. that the As are respectively = these equal

oms 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 A ABC = A EBC, 2. that A EBC = A DBC, i. e., less = greater, which

shows the supposition to be false, 3. that :. AD || BC.

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. Ś

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 A ARC = A GEF, 2. that A GEF = A DEF, i.e., less = greater, which

shows the supposition to be false, 3. that :: AD || BF.

PROPOSITION XLI.
E Theorem. If a parallelogram

and a triangle be upon the same
base, and between the same paral-
lels; the parallelogram shall be

double of the triangle. Proved by showing that cm BD = 2 A ABC, which = 2 A EBC.

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.

Steps of the Demonstration. 1. Prove that whole A ABC = 2 A ACE, 2. that Om FC = A ABC (and it has Z cer =

<D by construction.) PROPOSITION XLIII. Theorem. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.

Steps of the Demonstration. 1. Prove that A ABC = A ACD,

that A AEK + AKGC = A AKH + A KCF, that complement Bk = complement ko. that co

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 ZS BHF + HFE < 2 right Zs, and that consequently and FE will meet; and to state that FL is a om of which LB, BF are complements of me about the diameter hk.

Steps of the Demonstration. 1. Prove that om LB = A C,

that _ ABM = Z D.

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

KH M In the construction of the figure it is necessary to show,1. that _ FKH = L GHM,

in order 2. that ZS FKH + KHG = LS KHG + GHM, } to 3. that ZS KHG + GHM =2 right Zs, prove, 4. that ky, hm are in one right line.

(In a similar manner it must be proved that fg, Gl are in one right line.)

Also that fm is a Dm.

The Demonstration itself is merely the showing from the construction that the whole om = whole rectilineal figure, because the parts of the one = the parts of the other.

PROPOSITION XLVI.

Problem. To describe a square upon a given straight line.

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