"If a straight line meets two straight lines, so as to make the two " interior angles on the same side of it taken together less than " two right angles, these straight lines being continually produc"ed shall at length meet upon that fide on which are the angles " which are less than two right angles. See the notes on Prop. "29. of Book I," PROPOSITION I. PROBLEM. O describe an equilateral triangle upon a given fi- Let AB be the given straight line, it is required to describe an equilateral triangle upon it. From the center A, at the di stance AB describe the circle BCD. and from the center B, at the distance BA describe the circle ACE; and from the point Cin D A which the circles cut one another draw the straight lines b CA, CB to the points A, B. ABC shall be an equilateral triangle. C BE Book I. a. 3d Poftulate. b. ad Poft. Because the point A is the center of the circle BCD, AC is equal ' to AB. and because the point B is the center of the circle ACE, c. 15th DeBC is equal to BA. but it has been proved that CA is equal to finition. AB; therefore CA, CB are each of them equal to AB. but things which are equal to the fame are equal to one anotherd; therefore d. ist AxiCA is equal to CB. wherefore CA, AB, BC are equal to one another, and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB, Which was required to be done. F ROM PROP. II. PROB. a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC.. om. Book I. し e. 15. Def. f. 3. Ax. 2. 3. 1. 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 & BG. but it has been shewn that BC is equal to BG; wherefore AL and BC are each of them equal to BG. and things that are equal to the same 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. c. r. Ax. b. 3. Post. ftance AD describe the circle DEF. and because A is the center of the circle DEF, AE shall be equal to AD. but the straight line C is likewife equal to AD. whence AE and C are each of them equal to AD. wherefore the straight line AE is equal to oC, 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. PROP. IV. THEOREM. F two triangles have two fides of the one equal to two fides of the other, each to each; and have likewise the angles contained by those sides equal to one another: they shall likewife have their bases, or third fides, equal; and the two triangles shall be equal; and their other angles shall be equal, each to each, viz. those to which the equal fides are oppofite. Let ABC, DEF be two triangles which have the two fides AB, AC equal to the two fides DE, DF, each to each, viz. AB to DE, D Book I. and AC to DF; and the angle A BAC equal to the angle EDF. angle DEF, and the angle ACB to DFE. CE F For if the triangle ABC be applied to DEF so that the point A may be on D, and the straight line AB upon DE; the point B shall coincide with the point E, because AB is equal to DE. and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF, wherefore also the point C shall 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 base BC shall 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 base EF, two straight lines would inclose a space, which is impoffible. There- a. 10. Ax. fore the base BC shall coincide with the base 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 shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore 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 those sides equal to one another; their bases shall 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. T HE angles at the base of an Isosceles triangle are equal to one another; and if the equal sides be produced, the angles upon the other fide of the base shall be equal. : Let ABC be an Isosceles triangle, of which the side AB is equal Book I. to AC, and let the straight lines AB, AC be produced to Dand E. the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCСЕ. 2. 3. 1. b. 4. 1. In BD take any point F, and from AE, the greater, cut off AG equal to AF, the less, and join FC, GB. Because 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 B F C. 3. Аҳ. A C G E are equal; the remainder BF shall 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 opposite. therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. and since it has been demonftrated that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are also equal; the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the bafe 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. I COROLLARY. Hence every equilateral triangle is also equiangular. PROP. VI. THEOR. F two angles of a triangle be equal to one another, the fides also which subtend, or are opposite to, the equal angles shall be equal to one another. |