L E K H G F B the equilateral triangle D A B and let the right lines DA, DB be continued out to E and F [by poft. 2.]: then from the centre B, with the distance B C, describe a circle C G H [by poft. 3.]. And again, about the centre D, with the distance D G, let the circle G L K be defcribed. Therefore because the point B is the centre of the circle CGH, the right line B C will be equal to the right line B G [by def. 15.] Again, because the point D is the centre of the circle GLK; therefore DL is equal to DG [def. 15.]. But D A is equal to B [by conftr.] wherefore the remainder A L is equal to the remainder B G [by axiom 3.]. But it has been proved, that the right line BC is equal to the right line BG; therefore each of the right lines AL, BC, is equal to the right line BG: but things that are equal to the fame thing, are equal to one another; and therefore the right line A L is equal to the right line B C. Wherefore a right line AL is put at a given point A, equal to a given right line в C. Which was to be done. PROP. III. PROB L. Two unequal right lines being given, to take a right line from the greater equal to the leffer". 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 leffer. 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 diftance A D, let a circle D E F be described [by post. 3.]. And because 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 of 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 right line c. Therefore two unequal right lines, A B and C, being given, there is taken from the greater A B, a line equal to the leffer. Which was to be done. " The manual operation of this and the laft problem may be much shorter performed with a pair of companies, or other fuch like inftrument. But as thefe inftrumental operations are no poftulatums of Euclid's, they are therefore not to be admitted in the pure elements of geometry. Euclid himself, doubtless, very well knew that thefe problems might be mcchanically refolved more eafily by compaff.s; and even that he might have made poftulatums, 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 sight line. But he thought it not right fo to do, well knowing, that taking too many things for granted, not only leffens fcience, but may miflead, and beget error. And, on the contrary, the geometrical conftructions and demonftrations, even of propofitions of no great apparent value in therafelver, are always beft to be given, were it for no other reason, than to beget a habit of exactness in the geometrician, and thereby a ftronger difpofition in him to avoid error, and obtain truth. If two triangles have two fides of the one equal to two fides of the other, each to each; and bave one angle of the one equal to one angle of the ether, 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 equal to the other triangle; and the remaining angles of the one fall be equal to the remaining angles of the other, each to each, which are oppofite to the equal fidos w. Let the two triangles be A BC, DEF, having the two fides A B, A C of the one equal to the two fides DE, DF of the other, each to each, viz. A B to D E, and AC to DF; and the angle B A C of the one, equal to the angle EDF of the other. I fay, the bafe B c is equal to the bafe E F, and the triangle A B C fhall be equal to the triangle EDF; and the remaining angles fhall be equal to the remaining angles, each each to each, which are oppofite to the equal fides; that is, the angle ABC to the angle D E F, and the angle ACB to the angle DFE. B For if the triangle ABC be applied to the triangle E D F, CE D and the point A be put upon the point D, and the right line A B upon the right line D E, then fhall the point в agree with the point E; because A B is equal to DE: But fince F the right line A B agrees with D E, the right line A c fhall 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 в 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 neceffary that two right lines muft contain a space, which [by ax. 12.] is impoffible. Therefore the bafe BC agrees with the bafe EF, and accordingly is equal to it. Therefore the whole triangle ABC agrees with the whole triangle DEF, and will be equal to the fame; and the remaining angles will agree with the remaining angles, and therefore fhall 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 fides of the one equal to two fides 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 bafe of the one be equal to the base of the other, and one triangle fhall 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 oppofite 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 fenfual one made with the hands and the eyes; have taken occafion to find fault with the demonftration of this propofition, falfly affirming it to be only a mechanical demonstration, and not a geometrical one. PROP. PROP. V. THEOR. The angles at the bafe of ifofceles triangles are equal to one another; and the equal right lines being produced, the angles under the base shall be equal to one another x Let ABC be an ifofceles triangle, having the fide A B equal to the fide A c. And [by poft. 2.] continue the equal fides AB, AC directly forwards to D and E: I fay, 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 A G equal to AF the lefs [by prop. 3.] and draw the right lines FC, G B. D B E Therefore becaufe A F is equal to AG, and the right line A B equal to the right F line AC; there are two right lines FA, AC, equal to two right lines GA, AB, each to each, containing the common angle FAG: wherefore [by prop. 4.] the base F C will be equal to the base GB; and the triangle AFC equal to the triangle A G B and the remaining angles fhall be equal to the remaining angles, each to each, which are oppofite to the equal fides, viz. the angle ACF to the angle A B G, and the angle AFC to the angle AG B. But fince the whole line AF is equal to the whole line A G, and A B is equal to A C; 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 G B; therefore there are two right lines BF, FC, equal to two right lines CG, GB, each to each; and the angle B F C equal to the angle C GB; and the right line B c is their common bafe: wherefore the triangle B FC [by prop. 4.] fhall be equal to the triangle CGB; and the remaining angles fhall be equal to the remaining angles, each to each, which are oppofite to the equal fides: Therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle GBC. Now fince it has been demonftrated, that the whole angle A BG is equal to the whole angle AC F, and the angle c equal to to the angle B C F; the remaining angle A B C [by ax. 3.] will be equal to the remaining angle ACB; and they are at the bafe of the triangle A B C. But alfo it has been demonftrated, that the angle F B C is equal to the angle G C B, and they are under the base. Therefore the angles at the bafe of ifofceles triangles are equal to one another, and the equal right lines being produced, the angles under the bafe fhall be equal to one anWhich was to be demonftrated. other. * Scarborough fays, the latter part of this propofition feems not to be Euclid's, but put in by fomebody elfe; because, says he, it is certain, that Euclid never laid down an elementary propofition useless in any part thereof; nor never put that for one part of the propofition, which is only used as a means to prove the other. But herein I differ from Scarborough; for why might not Euclid himself demonftrate the latter part of his propofition firft? If it should be asked, Why then he did not fet it down firft? I answer, Because the latter part was the less important and useful part of the propofition, and therefore the former part deferved the first place. And if a propofition be not all Euclid's, because it is not useful in all its parts, other propofitions of thefe elements must not, in all their parts, be Euclid's; fuch as prop. 16. and 31. lib. 3. &c. But Scarborough is out here; for the latter part of this proposition is ufeful in the demonftration of one cafe of the 6th propofition; viz. when the point D falls within the triangle ACD; as alfo one cafe of the 24th propofition, when the point F (fee the figure of that propofition) falls within the triangle ED G. PROP. VI. THEOR. If two angles of a triangle be equal to one another, then fhall the fides oppofite to the equal angles be equal to one another. Let the triangle ABC have the angle ABC equal to the angle AC3. Ifay, the fide 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 leffer A c, and draw the right line D C. Then, |