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the equilateral triangle D A B,

and let the right lines D A, D B H

be continued out to E. and F D

[by post. 2.]: then from the C

centre B, with the distance B C,

describe a circle c G H [by poft. L

3.]. And again, about the E

centre D,

with the distance G

D G, let the circle G L K be T

defcribed. Therefore because the point E is the centre of the circle CGH, the right line B C will be equal to the right line BG [by def. 15.]. Again, because the point d is the centre of the circle 'GʻLK;. therefore D. L is equal to DG [def. 15.). But D A is equal to NB [by constr.] wherefore the remainder A L is equal to the remainder & G:[by axiom 3:]. But it has been proved, that the right line b c is equal to the right line BG; therefore each of the right lines A L, B C, is equal to the right line eg: but things that are equal to the same thing, are equal to one another; and therefore the right line A L is equal to the right line Bc.

Wherefore a right line Al is put at a given point A, equal to a given right line BC. Which was to be done.

!

PROP. III. PROBL.
Two unequal right lines being given, to take a right

line from the greater equal to the lefjer.'.
Let the two unequal right lines given be A B and c,
whereof let A B be the greater ; it is required to take a right
line from A B the greater, equal to c the lefser.

Put [by prop; 2.] at the point a a right line A D, equal to the right line c';" and from the centie A, with

the distance a D, let a circle DEF А. be described (by post. 3.]. And be

cause the point A is the centre of the circle D F E, the right line A E will be equal to the right line A D: but the right line c is equal to the

right line A D too. Therefore cach В

of

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of the right lines A E and c is equal to the right line Ad; wherefore the right line A E is alio equal to the righi line c.

Therefore two unequal right lires, A B and c, being given, there is taken from the greater A B, a line equal to c the lefser. Which was to be done.

• The manual operation of this and the las problem may be much shorter performed with a pair of companies, or other such like instrument. But as theło inftrumental operations are no poftulatums of Euclid's, they are ther-fore not to be almitted in the pure elements of geometry. Ewid himself, doubtless, very well knew that thele problems might be mcchanically resolved more easily by compafl s; ani even that he might have made poliulatums, to put a given right line at a given point equal to a given right line, and to cut off a right line from a given right line equal to a given i ht line. But he thought it not right fo to do, well knowing, that taking too many things for granted, not only left:ns scie..ce, but my miilead, and beget error. And, on the contrary, the geometrical constructions and demonstrations, even of proportions of no great apparent value in themselves, are always bift to l'e given, were it for no other reason, than to beget a hal it of exacineis in the geometrician, and thereby a stronger disposition in hiin to avoid error, and obtain truth,

PROP. IV. THEOR. If two triangles have two sides of the ore equal to

iwo sides of the other, each to each; and bate one angle of the one equal to cre aigle of the cher, viz. that which is contained 1.1.der the equal lins ; then spall the base of the one te equal to the base of the other ; ond 0e triangle equal to the other triangle; and the remaining angles of the one fall be equal to the remaining angles of the oth.r, each to each, which are opposite to the cqual sites w, Let the two triangles be A BC, D E F, having the two fides A B, A c of the one equal to the two lides Di, D F of the other, each to each, viz. A B to D I, and a C to DF; and the angle B A c of the on?, equal to the anzle E D F of the other. I say, the bale b c is equal to the base E F,

and the triangle A B C shall be equal to the triangle E DF; and the remaining angles ihall be equal to the remaining angles,

each

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each to each, which are opposite to the equal fides; that is, the angle ABC to the angle D E F, and the angle ACB to the angle DFE. For if the triangle A B C be applied to the triangle e DF,

D

and the point A be put upon the point D, and the right line A B upon the right line D E, then shall the point B agree with the point E; because A B

is equal to De: But since B CE

the right line A B agrees with D E, the right line A c shall also

A C shall also agree with DF, because the angle BAC is equal to the angle EDF. Wherefore the point c will agree with the point F, because the right line Ac is equal to the right line DF; but the point B will agree with the point E. And therefore the base Bc agrees with the base EF; for if the point B agrees with the point E, and the point c with the point F; but the base Bc does not agree with the base EF; it is necessary that two right lines must contain a space, which [by ax. 12.) is impossible. Therefore the base BC agrees with the base e F, and accordingly is equal to it. Therefore the whole triangle ABC agrees with the whole triangle D E F, and will be equal to the same; and the remaining angles will agree with the remaining angles, and therefore shall be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to the angle D F E.

If therefore two triangles have two sides of the one equal to two sides of the other, each to each; and have one angle of the one equal to one angle of the other, viz. that which is contained under the equal lines; then shall the base of the one be equal to the base of the other, and one triangle shall be equal to the other triangle ; and the remaining angles of the one shall be equal to the remaining angles of the other, each to each, which are opposite to the equal fides. Which was to be demonstrated.

w Some, for want of making a difference between a geometrical congruency and a mechanical one; that is, between an intellectual or mental one, and an actual sensual one made with the hands and the eyes; have taken occasion to find fault with the demonftration of this propofition, fallly affirming it to be only a mechanical demonitration, and not a geometrical one.

PROP.

PROP. V. THEOR. The angles at the base of isosceles triangles are equal

to one another; arid the equal right lines being produced, the angles under ihe baje sisall be equal to one another x

Let Aec be an isosceles triangle, having the side A B equal to the side Ac. And [by post. 2.) continue the equal fides A B, Ac directly forwards to D and E: I say, the angle ABC is equal to the angle ACB; and the angle CBD equal to the angle BCE.

For in the right line BD, let there be taken any point as F; and from the greater line A E, let be taken AG equal to AF the less [by prop. 3.) and draw the right lines FC, G B.

В. Therefore because a F is equal to AG, and the right line A B equal to the right ! line AC; there are two right lines FA, AC, equal to two right lines GA, A B, each to

D

E each, containing the common angle Fag: wherefore [by prop. 4.) the base F c will be equal to the base GB; and the triangle Arc equal to the triangle A G B6 and the remaining angles shall be equal to the remaining angles, each to each, which are opposite to the equal sides, viz. the angle Act to the angle A BG, and the angle A Fc to the angle A GB. But since the whole line AF is equal to the whole line A G, and A B is equal to AC; therefore the remainder BF will be equal to the remainder cg (by ax. 3.) But it has been proved, that the right line fc is equal to the right line GB; therefore there are two right lines B F, Fc, equal to two right lines CG, GB, each to each; and the angle brc equal to the angle c GB; and the right line bc is their common base : wherefore the triangle B F C [by prop. 4.) Ihall be equal to the triangle CGB; and the remaining angles shall be equal to the remaining angles, each to each, which are opposite to the equal fides : Therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle GBC. Now since it has been demonitrated, that the whole angle A BG is equal to the whole angle A C F, and the angle cec equal

to

to the angle BCF; the remaining angle A B C [by ax. 3.) will be equal to the remaining angle ACB; and they are at the base of the triangle A B C. But also it has been demonstrated, that the angle F B C is equal to the angle G CB, and they are under the base.

Therefore the angles at the base of i'osceles triangles are equal to one another, and the equal right lines being produced, the angles under the base shall be equal tò one another. Which was to be demonstrated.

* Scarborough fays, the latter part of this proposition seems not to be Euclid's, but put in by somebody elle; because, says he, it is certain, that Euclid never laid down an elementary proposition useless in any part thereof; nor never put that for one part of the proposition, which is only used as a means to prove the other. But herein I differ from Scarborough ; for why might not Euclid himself demonstrate the latter part of his propofition first? If it should be asked, Why then he did not set it down first? I answer, Because the latter part was the less important and useful part of the propofition, and therefore the fornier part deserved the first place. And if a proposition be not all Euclid's, because it is not useful in all its parts, other propofitions of these elements must not, in all their parts, be Euclid's; such as prop. 16. and 31. lib. 3. &c. But Scarborough is out here; for the latter part of this proposition is useful in the denonstration of one case of the 6th propofition; viz. when the point D falls within the triangle A CD; as alio one case of the 24th proposition, when the point p (see the figure of that proposition) falls within the triangle

EDG.

PROP. VI. THEOR.

If two angles of a triangle be equal to one another,

then mail the sides opposite to the equal angles be equal to one another.

Let the triangle A B C have the angle A bc equal to the angle A C3. I say, the side A c will be equal to the fide

A B.

For if ac be unequal to A B, one of them will be the greater. Let the greater be A B, and from the greater A B let be taken [by prop. 3.) the right line D B equal to the lesser A C, and draw the right line D c.

Then,

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