tion, as EG and FG; then, upon the same base EF, and upon the same side of it, there can be two triangles EDF, EGF, 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. PROP. IX. PROB. To bisect a given rectilineal angle, that is, to divide it into two equal angles. 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; but the base DF is also 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. SCHOLIUM. B D F E By the same construction, each of the halves BAF, CAF, may be divided into two equal parts; and thus, by successive subdivisions, a given angle may be divided into four equal parts, into eight, into sixteen, and so on. 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 common to the two triangles ACD, BCD: the two sides AC, CD, are equal to the two BC, CD, each to each; but the angle ACD is also 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 D. PROP. XI. PROB. To draw a straight line at right angles to a given straight line, from a given point in that line. 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. F 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 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 E B to each; but the base DF is also 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 (7. def.) 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. PROP. XII. PROB. To draw a straight line perpendicular to a given straight line, of an unlimited length, from a given point without it. Let AB be a given straight line, which may be produced to any length both ways, and let C be a point without it. It is required to draw a straight. line perpendicular to AB from the point C. C Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe (3. Post.) the circle EGF meeting AB in F, G: and bisect (10. 1.) FG in H, and join CF, CH, CG; the straight line CH, drawn from the given point C, is perpendicular to the given straight line AB. E A F Because FH is equal to HG, and HC common to the two triangles FHC, GHC, the two sides FH, HC are equal to the two GH, HC, each to each, but the base CF is also equal (11. Def. 1.) to the base CG; therefore the angle CHF is equal (8. 1.) to the angle CHG; and they are adjacent angles; now when a straight line standing on a straight line makes the adjacent angles equal to one another, each of them is a right angle, and the straight line which stands upon the other is called a perpendicular to it; therefore from the given point C a perpendicular CH has been drawn to the given straight line AB. PROP. XIII. THEOR. The angles which one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles. Let the straight line AB make with CD, upon one side of it the angles CBA, ABD; these are either two right angles, or are together equal to two right angles. For, if the angle CBA be equal to ABD, each of them is a right angle (Def. 7.); but, if not, from the point B draw BE at right angles (11. 1.) to CD; therefore the angles CBE, EBD are two right angles. Now, the angle CBE is equal to the two angles CBA, ABE together; add the angle EBD to each of these equals, and the two angles CBE, EBD will be equal (2. Ax.) to the three CBA, ABE, EBD. Again, the angle DBA is equal to the two angles DBE, EBA; add to each of these equals the angle ABC; then will the two angles DBA, ABC be equal to the three angles DBE, EBA, ABC; but the angles CBE, EBD have been demonstrated to be equal to the same three angles; and things that are equal to the same are equal (1. Ax.) to one another; therefore the angles CBE, EBD are equal to the angles DBA, ABC; but CBE, EBD, are two right angles; therefore DBA, ABC; are together equal to two right angles. COR. The sum of all the angles, formed on the same side of a straight line DC, is equal to two right angles; because their sum is equal to that of the two adjacent angles DBA, ABC. If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines are in one and the same straight line. At the point B in the straight line AB, let the two straight lines BC, BD upon the opposite sides of AB, make the adjacent angles ABC, ABD equal togethe. to two right angles. BD is in the same straight line with CB. For if BD be not in the same straight line with CB, let BE be in the same straight line with it; therefore, because the straight line AB makes angles with the straight line CBE, upon one side of A E C B D it, the angles ABC, ABE are together equal (13. 1.) to two right angles; but the angles ABC, ABD are likewise together equal to two right angles : therefore the angles CBA, ABE are equal to the angles CBA, ABD: Take away the common angle ABC, and the remaining angle ABE is equal (3. Ax.) to the remaining angle ABD, the less to the greater, which is impossible; therefore BE is not in the same straight line with BC. And in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which therefore is in the same straight line with CB. PROP. XV. THEOR. If two straight lines cut one another, the vertical, or opposite angles are equal. Let the two straight lines AB, CD, cut one another in the point E: the angle AEC shall be equal to the angle DEB, and CEB to AED. For the angles CEA, AED, which the straight line AE makes with the straight line CD, are together equal (13. 1.) to two right angles: and the angles AED, DEB, which the straight line DE makes with the straight line AB, are also together equal (13. 1.) to two right angles; therefore the two angles CEA, AED are equal to the two AED, DEB. Take away the common angle AED, and the remaining angle CEA is equal (3. Ax.) to the remaining angle DEB. In the same manner it may be demonstrated that the angles CEB, AED are equal. COR. 1. From this it is manifest, that if two straight lines cut one another, the angles which they make at the point of their intersection, are together equal to four right angles. COR. 2. And hence, all the angles made by any number of straight lines meeting in one point, are together equal to four right angles. PROP. XVI. THEOR. If one side of a triangle be produced, the exterior angle is greater than either of the interior, and opposite angles. Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite angles СВА, ВАС. Bisect (10. 1.) AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to G. Because AE is equal to EC, and BE to EF; AE, EB are equal to CE, EF, each to each; and the angle AEB is equal (15. 1.) to the angle CEF, because they are vertical angles; therefore the base AB is equal (4. 1.) to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore the angle ECD, that is ACD, is greater than BAE: In the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG, that is (15. 1.), the angle ACD, is greater than the angle ABC. 4 B E D PROP. XVII. THEOR. Any two angles of a triangle are together less than two right angles. Let ABC be any triangle; any two of its angles together are less than two right angles. Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is greater (16. 1.) than the interior and opposite angle ABC; to each of these add the angle ACB; therefore the angles ACD, ACB are greater than the angles ABC, ACB; but ACD, ACB are to B A C gether equal (13. 1.) to two right an D gles therefore the angles ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB as also CAB, ABC, are less than two right angles. PROP. XVIII. THEOR. The greater side of every triangle has the greater angle opposite to it. Let ABC be a triangle of which the side AC is greater than the side AB; the angle ABC is also greater than the angle BCA. From AC, which is greater than AB, cut off (3. 1.) AD equal to AB, and join BD: and because ADB is the exterior angle of the triangle BDC, it is greater (16. 1.) than the interior and opposite A D B |