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 impossible. (10. Ax.) Therefore 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 sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another, their bases shall likewise be equal, and the triangles be equal, and their other angles to which the equal sides are opposite shall be equal, each to each. Which was to be demonstrated. PROP. V. THEOR. THE 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 side of the base shall be equal. Let ABC be an isosceles triangle, of which the side AB is equal 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 (3. 1.) to AF, the less, and join FC, GB. A 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, AGB; therefore the base FC is equal (4. 1.) to the base GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are equal (4. 1.) to the remaining angles of the other, each to each, to which the equal sides are opposite; viz. the angle ACF to the angle ABG, and the angle AFC to the angle AGB; and because the whole AF is F equal to the whole AG, of which the parts AB, AC, are equal: the remainder BF shall be equal (3. Ax.) to the remainder D CG; and FC was proved to be equal to B G E GB; therefore the two sides 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 (4. 1.) and their remaining an BOOK I. THE ELEMENTS OF EUCLID. gles, each to each, to which the equal sides, 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 also 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. gular. Hence every equilateral triangle is also equian PROP. VI. THEOR. Ir two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another. Let ABC be a triangle having the angle ABC equal to the angle ACB; and the side AB is also equal to the side AC. D Α 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 (3. 1.) off DB equal to AC the less, and join DC; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB, each to each; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle (4. 1.) ACB, the less to the greater; which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles &c. Q. E. D. B Cor. Hence every equiangular triangle is also equilateral. PROP. VII. THEOR. C UPON the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. * If it be possible, let there be two triangles ACB, ADB, upon the same base AB, and upon the same side of it, which have their sides CA, DA, terminated in the extremity A of the base equal to one another, and likewise their sides, CB, DB, that are terminated in B. * See Note. Join CD; then, in the case 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 (5. 1.) 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 the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal (5. 1.) to the angle BCD; but A it has been demonstrated to be greater than it; which is impossible. C D E F Сс B 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 side of the base CD are equal (5. 1.) to one another, but the angle ECD is greater than the angle BCD; wherefore the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle BCD. Again because CB is equal to DB, the angle BDC is equal (5. 1.) to the angle BDC; but BCD has been proved to be greater than the same BCD ; A which is impossible. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. B Therefore upon the same base, and on the same side of it, there cannot be two triangles that have their sides 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. PROP. VIII. THEOR. Ir two triangles have two sides of the one equal to two sides. of the other, each to each, and have likewise their bases equal; the angle, which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them, of the other. For, if the triangle ABC be applied to DEF, so that the point B be on E, and the straight line BC upon EF; the point C shall also coincide with the point F. Because BC is equal to EF; therefore BC coinciding with EF, BA and AC shall coincide with ED and DF; for, if the base BC coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD but have a different situation, as EG, FG; then upon the same base EF, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity; but this is impossible; (7. 1.) therefore, if the base BC coincides 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 (8. Ax.) to it. Therefore, if two triangles, &c. Q. E. D. PROP. IX. PROB. To bisect a given rectilineal angle, that is, to divide it into two equal angles. A. Let BAC be the given rectilineal angle, it is required to bisect it. Take any point D in AB, and from AC cut (3. 1.) off AE equal to AD; join DE, and upon it describe (1. 1.) an equilateral triangle DEF; then join AF; the straight line AF bisects the angle BAC. Because AD is equal to AE, and AF is common to the two triangles DAF, EAF; the two sides DA, AF, are equal to the two sides, EA, AF each to each; and the base DF is equal to the base EF; therefore the angle DAF is equal (8. 1.) to the angle EAF; wherefore the given rectilineal angle BAC is bisected by the straight line, AF, which was B to be done. D E C PROP. X. PROB. To 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 (1. 1.) upon it an equilateral triangle ABC, and bisect (9. 1.) the angle ACB by the straight line CD. AB is cut into two equal parts in the point D. Because AC is equal to CB, and CD com. mon to the two triangles ACD, BCD; the two sides AC, CD are equal to BC, CD, each to each; and the angle ACD is equal to the angle BCD; therefore the base AD is equal to the base (4. 1.) DB and the straight line AB is divided into two equal parts in the point Which was to be done. D. PROP. XI. PROB. To draw a straight line at right angles to a given straight line, from a given point in the same. F 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 (3. 1.) make CE equal to CD, and upon DE describe (1. 1.) the equilateral triangle, DFE, and join FC; the straight line FC drawn from the given point C is at right angles to the given straight line AB. A D C E B Because DC is equal to CE, and FC common to the two triangles DCF, ECF; the two sides, DC, CF, are equal to the two EC, CF, each to each; and the base DF is equal to the base EF; therefore the angle DCF is equal (8. 1.) to the angle ECF; and they are adjacent angles. But, when the adjacent angles which one straight line makes with another straight line are equal to one another, each of them is called a right (10. Def. 1.) angle; therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given straight line AB, FC has been drawn at right angles to AB. Which was to be done. Cor. By help of this problem, it may be demonstrated, that two straight lines cannot have a common segment. If it be possible, let the two straight lines ABC, ABD have the segment AB common to both of them. From the point B draw BE at right angles to AB; and because ABC is a straight line, the angle CBE is equal (10. Def. 1.) to the angle EBA; in the same manner, because ABD is a straight line, the angle DBE is equal to the angle EBA; wherefore the angle DBE is equal to the angle CBE, the less to the greater; which is impossible; therefore two straight lines cannot have a common segment. E D A B C * See Note. |