2. 2. I. b. 3. Poft. c Because the point B is the center of the circle CGH, BC is equal to BG. and because D is the center of the circle GKL, DL is equal to DG, and DA, DB parts of them are equal; therefore the remainder AL is equal to the remainder f BG. but it has been fhewn that BC is equal to BG; wherefore AL and BC are each of them equal to BG, and things that are equal to the fame are equal to one another; therefore the straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. Which was to be done. FRO ROM the greater of two given ftraight lines to eut off a part equal to the lefs. Let AB and C be the two given ftraight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a parti equal to C the less. D E B F From the point A draw a the ftraight line AD equal to C; and from the center A, and at the dif tance AD defcribe the circle DEF. and because A is the center of the circle DEF, AE fhall be equal to AD. but the ftraight line C is likewise equal to AD. whence AE and C are each of c. I. Ax. them equal to AD. wherefore the straight line AE is equal to C, and from AB the greater of two ftraight lines, a part AE has been cut off equal to C the lefs. Which was to be done. IF two triangles have two fides of the one equal to two fides of the other, each to each; and have likewife the angles contained by thofe fides equal to one another: they fhall likewise have their bases, or third fides, equal; and the two triangles fhall be equal; and their other angles fhall be equal, each to each, viz. thofe to which the equal fides are opposite. Let ABC, DEF be two triangles which have the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AR to DE, and AC to DF; and the an- to each, viz. the angle ABC A CE to the angle DEF, and the angle ACB to DFE. For if the triangle ABC be applied to DEF fo that the point A may be on D, and the straight line AB upon DE; the point B fhall coincide with the point E, because AB is equal to DE. and AB coinciding with DE, AC fhall coincide with DF, because the angle BAC full on is equal to the angle EDF. wherefore alfo the point C fhall coincide with the point F, because the straight line AC is equal to DF., but the point B coincides with the point E; wherefore the bafe BC fhall coincide with the base EF. because the point B coinciding with E, and C with F, if the base BC does not coincide with the bafe EF, two ftraight lines would inclofe a space, which is impoffible. Therefore the bafe BC fhall coincide with the a. 10. Ax. bafe EF, and be equal to it. Wherefore the whole triangle ABC shall coincide with the whole triangle DEF, and be equal to it; and the other angles of the one, fhall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABCto the angle DEF, and the angle ACB to DFE.. Therefore if two triangles håve two sides of the one equal to two fides of the other, each to each, and have likewise the angles contained by thofe fides equal to one another; their bafes fhall likewife be equal, and the triangles be equal, and their other angles to which the equal fides are opposite, shall be equal, each to each. Which was to be demonstrated. THE PROP. V. THEOR. HE angles at the bafe of an Ifofceles triangle are equal to one another; and if the equal fides be produced, the angles upon the other fide of the base fhall be equal. Let ABC be an Ifofceles triangle, of which the fide AB is equal Book I. to AC, and let the ftraight lines AB, AC be produced to D and E. the angle ABC fhall be equal to the angle ACB, and the angle CBD to the angle BCE. 2. 3. I. b. 4. I. In BD take any point F, and from AE, the greater, cut off AG equal to AF, the lefs, and join FC, GB. Becaufe AF is equal to AG, and AB to AC; the two fides FA, AC are equal to the two GA, AB, each to each; and they contain the angle FAG common to the two tri angles AFC, AGB; therefore the base cause the whole AF is equal to the D B A G E c. 3. Ax. are equal; the remainder BF fhall be equal to the remainder CG. and FC was proved to be equal to GB; therefore the two fides BF, FC are equal to the two CG, GB, each to each; and the angle BFC is equal to the angle CGB; and the base BC is common to the two triangles BFC, CGB; wherefore the triangles are equal", and their remaining angles, each to each, to which the equal fides are oppofite. therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. and fince it has been demonftrated that the whole. angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are alfo equal; the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC. and it has also been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other fide of the base. Therefore the angles at the base, &c. ̧Q. E. D. COROLLARY. Hence every equilateral triangle is alfo equiangular. IF two angles of a triangle be equal to one another, the fides alfo which fubtend, or are oppofite to, the equal angles fhall be equal to one another. Let ABC be a triangle having the angle ABC equal to the Book I. angle ACB; the fide AB is alfo equal to the fide AC. For if AB be not equal to AC, one of them is greater than the other. let AB be the greater, and from it cut off DB equal to AC, a. 3. 1. the lefs, and join DC. therefore because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each; and the angle DBC is equal to the angle ACB; therefore the bafe DC is equal to the bafe AB, and the triangle DBC is equal to the triangle a ACB, the lefs to the greater; which B A D is abfurd. Fherefore AB is not unequal to AC, that is, it is UPO PROP. VII. THEOR. b. 4. I. PON the fame bafe, and on the fame fide of it, See N. there cannot be two triangles that have their fides which are terminated in one extremity of the bafe equal to one another, and likewife those which are terminated in the other extremity. If it be poffible, let there be two triangles ACB, ADB upon the fame base AB, and upon the fame fide of it, which have their fides CA, DA, terminated in the extremity A of the base, equal to one another, and likewife their fides CB, DB that are terminated in B. Join CD; then, in the cafe in which the Vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equal to the angle ADC. but the angle ACD is greater than the angle BCD, therefore the angle ADC is greater also than BCD; much more then is A B the angle BDC greater than the angle BCD. again, because CB is equal to DB, the angle BDC is equal to the angle BCD; but it has been demonstrated to be greater than it; which is impoffible. a. 5. I. Book I. a. 5. I. But if one of the Vertices, as D, be within the other triangle ACB; produce AC, AD to E, F. therefore because AC is equal to AD in the triangle ACD, the angles ECD, FDC upon the other fide of the base CD are equal a to one another; but the angle ECD is greater than the angle BCD, wherefore the angle FDC is likewife greater than BCD; much more then is the angle BDC greater than the angle BCD. again, because CB is eA qual to DB, the angle BDC is equal a to the angle BCD; but BDC has been proved to be greater than the fame BCD, which is impoffible. The cafe in which the Vertex of one triangle is upon a fide of the other, needs no demonstration. B Therefore upon the fame base, and on the fame fide of it, there cannot be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. Q. E. D. IF two triangles have two fides of the one equal to two fides of the other, each to each, and have likewife their bases equal; the angle which is contained by the two fides of the one fhall be equal to the angle contained by the two fides equal to them, of the other. Let ABC, DEF be two triangles having the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AB to DE, and AC to DF; and alfo the base BC equal A to the bafe EF. For if the trian gle ABC be applied D G to DEF fo that the B CE F point B be on E, and the straight line BC upon EF; the point C fhall alfo coincide with the point F, because BC is equal to EF. |