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angles ; and it has been shown that they cannot be greater ;
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
line, and on the same side of il, are between the same parallels.
Let the triangles ABC,
Join AD, and AE; the straight lines AD and BF shall be
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
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
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-
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
*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.
base, and between the same parallels ; the parallelogram shall
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.
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.
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.
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
is equal to the side EF, and the side AD to
Join BD, FH.
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.
Let ABCD, EFGH be two equal
The side AB of the one square, is equal to the side EF of the other square
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
other instance. Wherefore, universally, the sides of equal
greater than the square described upon a less.
A KB E • J. 3.
Make* AK, AI equal to EF, EH; and join KI. Join also
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,
And by parity of reasoning, the like may be proved in every
Let the square ABCD be greater than the square EFGH.
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
And by parity of reasoning, the