a 3. I. Book I. qual to AC, and let the straight lines AB, AC be produced to D and E, the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE. 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 sides FA, AC are equal to the two GA, AB, each to each ; and they contain the angle FAG common to the two triangles AFC, A AGB; therefore the base FC is eb. 4. I. qual to the base GB, and the tri- G E parts AB, AC are equal; the reC 3. Ax. mainder 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 bafe BC is common to the two triangles BFC, CGB; wherefore the triangles are equalb, 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 demonstrated, that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF, are allo 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 side of the base. Therefore the angles at the base, &c. Q. E. D. COROLLARY. Hence every equilateral triangle is also equiangular. PRO P. VI. THEOR. the sides also which subtend, or are opposite to, the equal angles shall be equal to one another. Le Let ABC be a triangle having the angle ABC equal to the Book I. angle ACB; the side , B is also equal to the side AĆ. For, if AB bę not equal to AC, one of them is greater than ibe other : Let AB be the greater, and from it cut off DB e- a 3. ti qual to AC, the less, and join DC; there A fore, because in the triangles DBC, ACB, DB is equal to AC, and EC common to D 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 bale DC is equal to the base AB, and the triangle DBC is equal to the triangle b ACB, the less to the b 4. 1 greater; which is absurd. Therefore B AB is not unequal to AC, that is, it is equal to it Wherefore, if two angles, &c. Q: E. D. Cor. Hence every equiangular triangle is also equilateral. U PRO P. VII. THEO R. there cannot be two triangles that have their fides If it be possible, let there be two triangles ACB, ADB, up- С D Join CD; then, in the case in á š. I. B But B 2 Book I. But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F; E F B Therefore upon the same base, and on the same fide of it, Q. E. D. PROP. VIII. THE OR. IF sides of the other, each to each, and have likewise Let ABC, DEF be two triangles having the two sides AB, D G For, if the tri. CE F BC BC is equal to EF; therefore BC coinciding with EF, BA and Book I. AC shall coincide with ED and DF; for, if the base BC coincides with the base EF, but the Gides BA, CA do not coincide with the fides ED, FD, but have a different situation, as EG, FG; then, upon the fame base EF, and upon the same fide of it, there can be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise their fides terminated in the other extremity : But this is impoffible; therefore if the base BC coin. a 7.1. cides with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal to it. Therefore b 8. Ax. if two triangles, &c. Q. E. D. T: po bifect a given rectilineal angle, that is, to divide it into two equal angles. Let BAC be the given rectilineal angle, it is required to bi. fect it. Take any point D in AB, and from AC cut off AE 2-8 3. . qual to ĄD; join DE, and upon iç describe an equilateral triangle DEF; А bi. I then join AF; the straight line AF bifects the angle BAC. Because AĎ is equal to AE, and AF is common to the two triangles D E DAF, EAF; the two sides DA, AF, are equal to the two fides EA, AF, each to each; and the base DF is e BI qual to the base EF; therefore the F с angle DAF is equal to the angle EXF; wherefore the given rectilineal angle BAC is bifected by the straight line AF. Which was to be done. C 8. I. Ti 10 bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line; it is required to divide it into two equal parts. Describe upon it an equilateral triangle ABC, and bisect * s. !. the angle ACB by the straight line CD. AB is cut into two o g. 1, equal parts in the point D. B 3 Because a 3: 1 b to TO straight line, from a given point in the fame, See N. Let AB be a given straight line, and C a point given in it; it is required to draw a straight line from the point C at right angles to AB. Take any point D in AC, and make CE equal to CD, and F the given Because DC is equal to CE, two EC, CF, each to each ; and the base DT is equal to the C 8. 1. bafe EF; therefore the angle DCF is equal to the angle ECF; which one straight line makes with another straight line are d 10. Def. equal to one another, each of them is called a right d angle ; therefore each of the angles DCF, ICF is a right angle. Cor. By help of this problem, it may be demonstrated, that If it be possible, let the two straight lines ABC, ABD have the fegment AB common to both of them. From the point B draw BE at right angles to AB; and because ABC is a straight line, E B 1. |