and (Constr.) 2. The base DF is equal to the base EF; therefore (I. 8.) wherefore 3. The angle DAF is equal to the angle EAF; 4. The angle BAC is bisected by the straight line AF. Which was to be done. PROP. X.-PROBLEM. 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 (I. 1.) upon it an equilateral triangle ABC, and bisect (I. 9.) the angle 4CB by the straight line CD. AB is cut into two equal parts in the point D. C Because AC is equal to CB, and CD common to the two triangles ACD, BCD; 1. and (Constr.) The two sides AC, CD are equal to BC, CD, each to each; 2. The angle ACD is equal to the angle BCD; therefore (I. 4.) 3. wherefore The base AD is equal to the base DB; 4. The straight line AB is divided into two equal parts in the point D. Which was to be done. PROP. XI. - PROBLEM. To draw a straight line at right angles to a given straight line, from a given point in the same. 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 (I. 3.) make CE equal to CD, and upon DE describe (I. 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 F Because DC is equal to CE, and FC common to the two triangles DCF, ECF; 1. The two sides DC, CF, are equal to the two EC, CF, each to each; and (Constr.) 2. therefore (I. 8.) 3. The base DF is equal to the base EF; The angle DCF is equal 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 (Def. 10.) a right angle; therefore, 4. Each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given straight line AB, 5. 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 becanse ABC is a straight line, (Def. 10.) 1. The angle CBE is equal to the angle EBA; in the same manner, because ABD is a straight line, 2. The angle DBE is equal to the angle EBA; wherefore (Ax. 1.) 3. The angle DBE is equal to the angle CBE, the less to the greater, which is impossible; therefore 4. Two straight lines cannot have a common segment. PROP. XII.-PROBLEM. To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. Let AB be the 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. Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe (Post. 3.) the circle EGF meeting AB in F, G; and bisect (I. 10.) FG in H, and join CH; the straight line CH, drawn from the given point C, is perpendicular to the given straight line AB. C Join CF, CG; and because FH is equal to HG, and HC common to the two triangles FHC, GHC, 1. The two sides FH, HC, are equal to the two GH, HC, each to each; and (Def. 15.) 2. The base CF is equal to the base CG; therefore (I. 8.) 3. The angle CHF is equal to the angle CHG; and they are adjacent angles; but when a straight line standing on another 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 (Def. 10.); therefore from the given point C 4. A perpendicular CH has been drawn to the given straight line AB. Which was to be done. PROP. XIII.-THEOREM. 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. E For, if the angle CBA be equal to ABD, each of them is a right angle (Def. 10.). But if not, from the point B draw BE at right angles (I. 11.) to CD; therefore (Def. 10.) 1. The angles CBE, EBD, are two right angles; and because CBE is equal to the two angles CBA, ABE, together, add the angle EBD to each of these equals; therefore (Ax. 2.) 2. The angles CBE, EBD, are equal to the three angles CBA, ABE, EBĎ. Again, because the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC; therefore (Ax. 2.) 3. The angles DBA, ABC, are 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 (Ax. 1.) to one another; therefore 4. The angles CBE, EBD, are equal to the angles DBA, ABC; but CBE, EBD, are two right angles; therefore (Ax. 1.) 5. DBA, ABC, are together equal to two right angles. Wherefore, the angles which one straight line, &c. Q.e.d. PROP. XIV. THEOREM. 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 shall be 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 together to two right angles. BD is in the same straight line with CB. A 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 it, (I. 13.) 1. The angles ABC, ABE, are together equal to two right angles; but the angles ABC, ABD are likewise together equal to two right angles (Hyp.); therefore (Ax. 1.) 2. The angles CBA, ABE, are equal to the angles CBA, ABD. Take away the common angle ABC, and (Ax. 3.) 3. The remaining angle ABE is equal to the remaining angle ABD, the less to the greater, which is impossible; therefore 4. 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, therefore 5. BD is in the same straight line with CB. Wherefore, if at a point, &c. Q.E.D. PROP. XV.-THEOREM. If two straight lines cut one another, the vertical, or opposite, angles shall be 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 AÊD. Because the straight line AE makes with CD the angles CEA, AED, (I. 13.) 1. CEA, AED, are together equal to two right angles. Again, because the straight line DE makes with AB the angles AED, DEB, (I. 13.) 2. AED, DEB, are together equal to two right angles; and CEA, AED, have been demonstrated to be equal to two right angles; wherefore (Ax. 1.) DEB. 3. The angles CEA, AED, are equal to the angles AED, |