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angles ; and it has been shown that they cannot be greater ;
wherefore ADC and DCB cannot be less. And because the
angles ADC and DCB together are neither greater nor less than
two right angles, they are equal to two right angles. But because

the angles ADC, DCB are together equal to two right angles, •1.27.Cor.2. AD and BC are* parallel.

And by parity of reasoning, the like may be proved of all
other triangles under the same conditions. Wherefore, univer-
sally, equal triangles upon the same base, &c. Which was to be
demonstrated.

PROPOSITION XL.
See Note. THEOREM.- Equal triangles, upon equal bases in the same straight

line, and on the same side of il, are between the same parallels.

Let the triangles ABC,
EFD, which are upon equal

н.
bases BC and EF in the
same straight line BF, and

E
on the same side of it, be
equal to one another. They are between the same parallels.

Join AD, and AE; the straight lines AD and BF shall be
parallel.

For if the angles DAB, ABF are not together equal to two

right angles, let it be assumed that they are greater ; and + 1. 28. from the point A drawt AG parallel to BF. Because AG, $1.29.Cor.1. BF are parallel, the angles GAB, ABF must be † together equal

to two right angles; and because the angles DAB, ABF are
supposed to be together greater than two right angles, they must

be* together greater than the angles GAB, ABF; and taking +INTERC. 1.

away the common angle ABF, the remaining angle DAB must bet Cor. 8.

greater than the remaining angle GAB. Wherefore the straight
line AG must lie within the straight line AD; and moreover it
must lie on the side of AE which is towards AD, for if not, it

would meet BE, which cannot be, for they are parallel ; there11.13.Cor.5. fore being prolonged it mustf meet DE, in some point bea

tween D and E. Let it meet it in H; and join HF. Be-
cause the triangles ABC, EFH are upon equal bases BC

and EF, and between the same parallels BF and AH, the *1. 38. triangle EFH must be* equal to the triangle ABC. But the Hyp) triangle EFD ist equal to the triangle ABC; therefore the

Cor. 2.

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*Interc. 1. triangle EFD must be* equal to the triangle EFH, the greater +I. Nom.26, to the less, which is impossible. The assumptions, ther

re, which involves this impossible consequence, cannot be true; or DAB, ABF are not together greater than two right angles. And in the same way may be shown that ADF, DFB cannot be together

greater than two right angles. Also they cannot be less ; for 11.32.Cor.1. because the interior angles of the quadrilateral figure ABFD are

together equal to four right angles, if ADF, DFB are together less than two right angles, the remaining angles DAB, ABF must be together greater than two right angles; and it has been shown that they cannot be greater ; wherefore ADF and DFB cannot be less. And because the angles ADF and DFB together are neither greater nor less than two right angles, they are

equal to two right angles. But because the angles ADF, DFB *1.27.Cor.2. are together equal to two right angles, AD and BF are* parallel.

[In the same way might the triangles have been shown to be between the same parallels, if the point E in which is terminated a side of the triangle EFD, had been situate within the base BC of the other triangle, or had coincided with the point C; or if the point D had coincided with the point A, or been on the other side of it.]

And by parity of reasoning, the like may be proved of all other triangles under the same conditions. Wherefore, universally, equal triangles, upon equal bases &c. Which was to be demonstrated.

PROPOSITION XLI.
THEOREM.-If a parallelogram and triangle be upon the same

base, and between the same parallels ; the parallelogram shall
be double of the triangle.

E

Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE; the parallelogram ABCD is double of the triangle EBC.

Join AC.

The triangle ABC is equal to half of the parallelogram +1, 33,

ABCD, because the diagonal AC+ bisects it; and the triangle 11.37.

ABC ist equal to the triangle EBC, because they are on the same

base BC and between the same parallels ; wherefore half the *INTERC. 1. parallelogram ABCD is* equal to the triangle ECB, and the

*INTERC. 1. parallelogram ABCD is equal to the double of the triangle Cor. 10.

EBC.

And by parity of reasoning, the like may be proved of every other parallelogram and triangle under the same conditions. Wherefore, universally, if a parallelogram and triangle be upon the same base, &c. Which was to be demonstrated.

SCROLIUM. In a similar manner might be demonstrated, that if a paral. lelogram and triangle be upon equal bases and between the same parallels,

the parallelogram shall be double of the triangle. See Note.

Cor. Every triangle is equal to half the rectangle made by any one of its sides, and its altitude as measured to such side or to its prolongation.

For such a rectangle is a parallelogram upon the same base, and between the same parallels. Therefore (by the Prop. above) it is the double of the triangle.

D

+1. 10.

#1.3.

PROPOSITION XLII.
PROBLEM.—To describe a square upon a given straight line.
Let AB be the given straight line; it is required

с to describe a square upon it.

From the points A and B, drawt AD and BC at right angles to AB and on the same side of it; and make$ AD and BC, each equal to AB; and join

А B CD. ABCD is the square required.

Because the angles BAD and ABC are two right angles, AD •1.27.Cor.2. and BC are* parallel. But they aret also equal; wherefore DC Constr. 11.34.

and AB which join their extremities, are equal and parallel, and 1. 32 bis. ABCD is* a parallelogram. And because the opposite sides are

equal to one another, and also AB equal to AD; all the four sides

are equal to one another. Also because one of the angles, as BAD, + 1. 3+ bis. is a right angle, all the angles aret right angles. Wherefore ABCD 11.34 bis. isf a square ; and it is described upon AB. Which was to be done. Nom. 2. And by parity of reasoning, the like may be done in every other instance.

PROPOSITION XLIII. THEOREM.Rectangles which are made by equal straight lines, See Note.

are equal to one another. Let ABCD, EFGH be two rectangles, of which the side AB

Nom.

D

G

is equal to the side EF, and the side AD to

с ня
the side EH. The rectangles are equal to one
another.

Join BD, FH.
• Hyp:
Because AB, AD are* equal to EF, EH

A в Е +1.11. respectively, and the angle BAD ist equal to 11.4. the angle FEH, the triangle BAD ist equal to the triangle FEH. * 1. 33.

And because a diagonal of any parallelogram divides* it into two equal parts, the rectangles ABCD, EFGH are respectively double

of the triangles BAD, FEH; and because the triangles are equal *INTERC. 1. to one another, the rectangles which are the double of them aret Cor. 10.

equal to one another.

And by parity of reasoning, the like may be proved in every other instance. Wherefore, universally, rectangles which are made by equal straight lines, &c. Which was to be demonstrated. COR.

The squares described upon equal straight lines, are equal to one another.

For such squares are rectangles made by equal straight lines.

с

H

G

D
I!

A

PROPOSITION XLIV.
THEOREM.The sides of equal squares, are equal to one another.

Let ABCD, EFGH be two equal
squares.

The side AB of the one square, is equal to the side EF of the other square

KB

F For if AB be not equal to EF, one of 1.3.

them must be the greater. Let AB be the greater; and make AK and AI each equal to EF ; and join Kl. Join also BD, FH.

Because AK, AI are equal to EF, EH respectively, and the *1.11. I. 4. angle KAI is* equal to the angle FEH, the triangle KAI must bet

equal to the triangle FEH. But because the triangles BAD, 31. 33.

FEH are the halves of the equal squares ABCD, EFGH, they *INTERC. 1. are* equal to one another ; therefore the triangle KAI (which is + INTERC. 1. equal to the triangle FEH) must bet equal to the triangle BAD,

the less to the greater, which is impossible. Therefore AB is not greater than EF; and in like manner may be shown that EF is not greater than AB. And because neither is greater than the other, they are equal.

And by parity of reasoning, the like may be proved in every

1

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other instance. Wherefore, universally, the sides of equal
squares, &c. Which was to be demonstrated.

PROPOSITION XLV.
THEOREM.—The square described upon a greater straight line, is

greater than the square described upon a less.
Let the straight line AB be greater D
than the straight line EF. The square

I

H
ABCD which is described upon AB,
is greater than the square EFGH
which is described

upon
EF.

A KB E • J. 3.

Make* AK, AI equal to EF, EH; and join KI. Join also
BD, FH.

Because AK, AI are equal to EF, EH respectively, and the +1.11. angle KAI ist equal to FEH, the triangle KAI ist equal to the 11.4.

triangle FEH. But the triangle BAD is greater than the triangle "Interc. 1. KAI; therefore it is* also greater than the triangle FEH, Cor. 2. + 1. 33. Wherefore the square ABCD which ist double of the triangle INTERC. 1. BAD, ist greater than the square EFGH which is double of the Cor. 11,

triangle FEH.

And by parity of reasoning, the like may be proved in every
other instance. Wherefore, universally, the square described upon
a greater straight line, &c. Which was to be demonstrated.

PROPOSITION XLVI.
See Note, THEOREM.—The greater square has the greater side.

Let the square ABCD be greater than the square EFGH.
The side AB is greater than the side EF.

For if it be not greater, it must either be equal or less. It is not *1, 43. Cor, equal; for then the square ABCD must be* equal to the square + Hyp. EFGH; but ABCD is not equal to EFGH, for it ist greater. * I. 45.

Neither is it less; for then the square ABCD must bef less
than the square EFGH; but ABCD is not less, for it is greater.
Wherefore, because the side AB is D
neither equal to the side EF, nor

I

H
less, it is greater.

And by parity of reasoning, the
like may be proved in every other A KB
instance. Wherefore, universally, the greater square &c. Which
pyas to be demonstrated,

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