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and AG, ME, are the parallelograms about HK;
also LB, BF are the complements ; therefore the complement LB is equal to the complement BF; (1. 43.) but the complement BF is equal to the triangle C; (constr.)
wherefore LB is equal to the triangle C. And because the angle GBE is equal to the angle ABM, (1. 15.) and likewise to the angle D; (constr.).
therefore the angle ABM is equal to the angle D. (ax. 1.) Therefore to the given straight line AB, the parallelogram LB has een applied, equal to the triangle C, and having the angle ABM ual to the given angle D. Q.E.F.
PROPOSITION XLV. PROBLEM, To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.
Let ABCD be the given rectilineal figure, and E the given rectilineal angle.
It is required to describe a parallelogram that shall be equal to the figure AB CD, and having an angle equal to the given angle E.
Join DB. Describe the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to the angle E; (1. 42.) to the straight line G#, apply the parallelogram GM equal to the
triangle DBC, having the angle GHM equal to the angle E.
(I. 44.) Then the figure FKML shall be the parallelogram required. Because each of the angles FKH, GHM, is equal to the angle E, therefore the angle FKH is equal to the angle GHM;
add to each of these equals the angle ŘHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM;
but FK7, KHG are equal to two right angles ; (I. 29.) therefore also KHG, GHM are equal to two right angles ;
and because at the point H, in the straight line GĦ, the two straight lines KH, HM, upon the opposite sides of it, make the adjacent angles KHG, GHM equal to two right angles,
therefore HK is in the same straight line with HM. (1. 14.)
And because the line HG meets the parallels KM, FG, therefore the angle MHG is equal to the alternate angle HGF; (I. 29.)
add to each of these equals the angle HGL; therefore the angles MHG, HGL are equal to the angles HGF, HGL; but the angles MHG, HGL are equal to two right angles; (I. 29.) therefore also the angles HGF, HGL are equal to two right angles,
and therefore FG is in the same straight line with GL. (I. 14.)
And because KF is parallel to HG, and HG to ML,
therefore KF is parallel to ML; (I. 30.).
and FL has been proved parallel to KM,
wherefore the figure FKML is a parallelogram; and since the parallelogram HF is equal to the triangle ABD,
and the parallelogram GM to the triangle BDC; therefore the whole parallelogram KFLM is equal to the whole
rectilineal figure ABCD. Therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Q.E.F.
COR. From this it is manifest how, to a given straight line, to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure; viz. by applying to the given straight line a parallelogram equal to the first triangle ABD, (1. 44.) and having an angle equal to the given angle.
It is required to describe a square upon AB.
make AD equal to AB; (1. 3.)
therefore ABÈD is a parallelogram ;
but AD is equal to AB, therefore the four lines AB, BE, ED, DA are equal to one another
and the parallelogram ABED is equilateral.
since AD meets the parallels AB, DE, therefore the angles BAD, ADE are equal to two right angles ; (1.29.)
but BAD is a right angle; (constr.)
therefore also ADE is a right angle. But the opposite angles of parallelograms are equal;. (1. 34.) therefore each of the opposite angles ABE, BED is a right angle;
wherefore the figure ABED is rectangular,
and it has been proved to be equilateral ;
COR. Hence, every parallelogram that has one of its angles a right angle, has all its angles right angles.
PROPOSITION XLVII. THEOREM. In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.
Let ABC be a right-angled triangle, having the right angle BAC!
Then the square described upon the side BC, shall be equal to the squares described upon BA, AC.
On BC describe the square BDEC, (I. 46.)
and on BA, AC the squares GB, HC;
and join AD, FC.
and that the angle BAG is a right angle, (def. 30.) the two straight lines AC, AG upon the opposite sides of AB, make with it at the point A, the adjacent angles equal to two right angles ;
therefore CA is in the same straight line with AG. (1. 14.) For the same reason, B A and AH are in the same straight line. And because the angle DB C is equal to the angle FBA,
each of them being a right angle,
add to each of these equals the angle ABC, therefore the whole angle ABD is equal to the whole angle FBC. (ax.2.)
And because the two sides AB, BD, are equal to the two sides FB, BC, each to each, and the included angle ABD is equal to the included angle FBC,
therefore the base AD is equal to the base FC, (I. 4.)
and the triangle ABD to the triangle FBC. Now the parallelogram BL is double of the triangle ABD, (1. 41.) because they are upon the same base BD, and between the same parallels BD, AL;
also the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the
same parallels FB, GĊ.
Similarly, by joining AE, BK, it can be proved,
Therefore the whole square BDEC is equal to the two squares GB, HC; (ax. 2.) and the square BDEC is described upon the straight line BC,
and the squares GB, HC, upon AB, AC: therefore the square upon the side BC, is equal to the squares upon the sides AB, AC.
Therefore, in any right-angled triangle, &c. Q. E.D.
PROPOSITION XLVIII. THEOREM. If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it; the angle contained by these two sides is a right angle. Let the
square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other two sides, AB, AC.
Then the angle BAC shall be a right angle.
From the point A draw AD at right angles to AC, (1. 11.)
make AD equal to AB, and join DC.
Then, because AD is equal to AB,
to each of these equals add the square on AC; therefore the squares on AD, A Care equal to the squares on AB, AC: but the squares on AD, AC are equal to the square on DC, (I. 47.)
because the angle DAC is a right angle; and the square on BC, by hypothesis, is equal to the squares on BA, AC;
therefore the square on DC is equal to the square on BC;
and therefore the side DC is equal to the side BC.
and AC is common to the two triangles DAC, BAC; the two sides DA, AC, are equal to the two BA, AC, each to each ; and the base DC has been proved to be equal to the base BC; therefore the angle DAC is equal to the angle BAC; (I. 8.)
but DAC is a right angle;
therefore also BAC is a right angle. Therefore, if the square described upon, &c. Q. E. D.
ON THE DEFINITIONS. GEOMETRY is one of the most perfect of the deductive Sciences, and seems to rest on the simplest inductions from experience and observation.
The first principles of Geometry are therefore in this view consistent hypotheses founded on facts cognizable by the senses, and it is a subject of primary importance to draw a distinction between the conception of things and the things themselves. These hypotheses do not involve any property contrary to the real nature of the things, and consequently cannot be regarded as arbitrary, but in certain respects, agree with the conceptions which the things themselves suggest to the mind through the medium of the senses. The essential definitions of Geometry therefore being inductions from observation and experience, rest ultimately on the evidence of the senses.
It is by experience we become acquainted with the existence of individual forms of magnitudes; but by the mental process of abstraction, which begins with a particular instance, and proceeds to the general idea of all objects of the same kind, we attain to the general conception of those forms which come under the same general idea.
The essential definitions of Geometry express generalized conceptions of real existences in their most perfect ideal forms: the laws and appearances of nature, and the operations of the human intellect being supposed uniform and consistent.
But in cases where the subject falls under the class of simple ideas, the terms of the definitions so called, are no more than merely equivalent expressions. The simple idea described by a proper term or terms, does not in fact admit of definition properly so called. The definitions in Euclid's Elements may be divided into two classes, those which merely explain the meaning of the terms employed, and those, which, besides explaining the meaning of the terms, suppose the existence of the things described in the definitions.
Definitions in Geometry cannot be of such a form as to explain the nature and properties of the figures defined : it is sufficient that they give marks whereby the thing defined may be distinguished from every other of the same kind. It will at once be obvious, that the definitions of Geometry, one of the pure sciences, being abstractions of space, are not like the definitions in any one of the physical sciences. The discovery of any new physical facts may render necessary some alteration or modification in the definitions of the latter.
Def. 1. Simson has adopted Theon's definition of a point. Euclid's definition is, onuelov &OTIV où Mépos oudév, “A point is that, of which there is no part, or which cannot be parted or divided, as it is explained by Proclus. The Greek term onuecov, literally means, a visible sign or mark on a surface, in other words, a physical point. The English term point, means the sharp end of any thing, or a mark made by it. The word point comes from the Latin punctum, through the French word point. Neither of these terms, in its literal sense, appears to give a very exact notion of what is to be understood by a point in Geometry. Euclid's definition of a point merely expresses a negative property, which excludes the proper and literal meaning of the Greek term, as applied to denote a physical point, or a mark which is visible to the senses.
Pythagoras defined a point to be μονας θέσιν έχουσα, monad having position. By uniting the positive idea of position, with the negative idea of defect of magnitude, the conception of a point in Geometry may