If 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 is equal to the angle contained by the two sides equal to them, of the other. Let ABC, DEF be_two_triangles, having the two sides AB, AC equal to the two sides DE, DF, each to each; viz., AB to DE, and AC to DF; and also the base BC equal to the base EF. The angle BAC is equal to the A angle EDF. B D G CE For, if the triangle ABC be applied to the triangle DEF, so that the point B shall be on E, and the base BC upon the base EF; then, the point C shall coincide with the point F, because BC is equal (Hyp.) to EF. And BC coinciding with EF, BA and AC shall coincide with ED and DF. For, if the base BC coincides with the base EF, and the sides BA, AC do not coincide with the sides ED, DF, but have a different situation as EG, GF. Then, upon the same base EF, and upon the same side of it, there can be two triangles having their sides terminated in one extremity of the base equal to one another, and likewise those terminated in the other extremity. But this is (I. 7) impossible._Wherefore, if the base BC coincides with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF. Therefore, the angle BAC coincides with the angle EDF, and is equal (4x. 8) to it. Therefore if two triangles, &c. Q. E. D. Corollary. It is a plain inference from the demonstration of this proposition, that if the three sides of one triangle be equal to the three sides of another triangle, each to each; the three angles of the one shall be equal to the three angles of the other, each to each,-viz., those to which the equal sides are opposite. Pappus demonstrates this proposition without the aid of Prop. VII., by applying the base of the one triangle to the base of the other, and inverting the former, so that they shall correspond exactly to the terms of the proposition contained in the exercise appended to the preceding proposition. To bisect a given rectilineal angle; that is, to divide it into two equal angles. AD. Because AD is equal (Const.) 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 (Const.) to the base EF. Therefore the angle DAF is equal (I. 8) to the angle EAF. Wherefore the given rectilineal angle B BAC is bisected by the straight line AF. Q. E. F. T By means of this proposition, an angle may be divided into any number of equal parts denoted by the successive powers of the number 2; that is, into 2, 4, 8, 16, 32, 64, &c., equal parts; but to divide any angle into three equal parts, that is, to trisect an angle, is beyond the power of elementary geometry. 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 upon AB (I. 1) an equilateral triangle ABC, and bisect (I. 9) the angle ACB by the straight line CD, meeting A B in the point D. Then, AB is divided into two equal parts at the point D. Because AC is equal (Const.) to CB, and CD common to the two triangles ACD, BCD, the two sides AC, CD, are equal to the two sides B C, CD, each to each; and the angle ACD is equal (Const.) to the angle BCD. Therefore the base AD is equal to the base (I. 4) DB, and the straight line AB is divided into two equal parts at the point D. Q.E. F. By this problem a straight line may be divided into any number of equal parts denoted by the series 2, 4, 8, 16, &c. To draw a straight line at right angles to a given straight line, from a given point in the same. Let A B be a given straight line, and C a given point 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 (I. 3) CE equal to ČD. 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. B Because DC is equal (Const.) to CE, and FC common to the two triangles DCF, ECF, the two sides DC, CF, are equal to the two sides EC, CF, each to each; and the base DF is equal (Const.) to the base E F. Therefore the angle DCF is equal (I. 8) 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 (Def. 10) angle. Therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given straight line AB, a straight line FC has been drawn at right angles to AB. Q. E. F. COR.-Two straight lines cannot have a common segment; that is, they cannot coincide in part without coinciding altogether. 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 (J. 11) BE at right angles to A B. Because ABC is a straight line, the angle CBE is equal (Def. 10) to the angle EBA. Because ABD is a straight line, the angle DBE is equal to the angle EBA. Therefore (Ax. 1) 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. Q. E.D. B D In Playfair's edition of the Elements, this corollary is drawn from the defective definition of a straight line, which he has substituted for Euclid's. Exercises.-1. From the extremity of a given straight line, to draw a straight line which shall make a right angle with it. 2. In a straight line of unlimited length, given in position, to find a point equally distant from two given points. 3. If the three sides of a triangle be bisected, and straight lines be drawn through the points of bisection at right angles to the sides, they shall, if produced, meet in the same point. 4. To describe a square upon a given straight line. To draw a straight line perpendicular to a given straight line of 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 and G. 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. Join CF, CG. Because FH is equal (Const.) to HG, and HC common to the two triangles FHC, GH C, the two sides FH, HC, are equal to the two sides G H, EC, each to each; and the base CF is equal (Def. 15) to the base CG. Therefore the angle CHF is equal (I. 8) 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 (Def. 10) to it. Therefore from the given point C a perpendicular CH has been drawn to the given straight line AB. Q. E.D. Exercise. From a point nearly in the perpendicular to a straight line at one extremity, but within the right angle, to draw a straight line which shall make a right angle with it. 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 two angles CBA, ABD. These angles are either two right angles, or are together equal to two right angles. For if the angle CBA be equal to the angle ABD, each of them is a right angle (Def. 10). But if the angles CBA, ABD, be unequal, from the point B draw B E at right angles (I. 11) to CD. Therefore the angles CBE, EBD (Def. 10) are two right angles. But the angle CBE is equal to the two angles CBA, ABE, together. To each of these equals add the angle EBD. D D Therefore the two angles CBE, EBD, are equal (Ax. 2) to the three angles CBA, ABE, EBD. Again, the angle D BA is equal to the two angles DBE, EBA. To each of these equals add the angle A B C. Therefore the two angles DBA, A B C, are equal (Ax. 2) to the three angles DBE, EBA, ABC. But the two angles CB E, EBD, have been proved to be equal to the same three angles. And things that are equal to the same thing are equal (Ax. 1) to one another. Therefore the two angles CBE, EBD, are equal to the two angles DBA, AB C. But the two angles CBE, EBD, are two right angles. Therefore the two angles D B A, ABC, are together equal (Ax. 1) to two right angles. Wherefore, the angles which one straight line, &c. Q. E.D. Corollary 1.-All the angles made by any number of straight lines meeting at a point, on one side of a straight line, are together equal to two right angles. Corollary 2.-All the angles made by any number of straight lines meeting in a point, are together equal to four right angles. Definition 1.-When two angles are together equal to two right angles, the one is called the supplement of the other. Definition 2.-When two angles are together equal to a right angle, the one is called the complement of the other. Exercise. If an angle and its supplement be bisected, the bisecting lines are perpendicular to each other. PROP. XIV. THEOREM. If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the two 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 together equal to two right angles. Then, BD is in the same straight line with CB. C B For, if BD be not in the same straight line with CB, let BE be in the same straight line with it. Because the straight line AB makes with the straight line CBE, upon one side of it, the two angles ABC, ABE, these two angles are together equal (I. 13) to two right angles. But the two angles ABC, ABD are likewise together equal (Hyp.) to two right angles. Therefore the two angles CBA, ABE are equal (Ax. 1) to the two angles CBA, ABD. From each of these equals, take away the common angle ABC. Therefore the remaining angle ABE is equal (Ax. 3) to the remaining angle ABD, the less to the greater, which is impossible. Wherefore BE is not in the same straight line with BC. In like manner, it may be shown that no other straight line but BD, can be in the same straight line with B C. Therefore BD is in the same straight line with CB. Wherefore, if at a point, &c. Q. E. D. Corollary. If at a point in a straight line, two other straight lines upon the same side of it, make each a right angle with it, these two straight lines shall coincide with each other. PROP. XV. THEOREM. If two straight lines cut one another, the vertical, or opposite angles are equal. Let the two straight lines A B, CD, cut one another in the point E. The angle AEC is equal to the angle DEB, and the angle CEB to the angle AED. Because the straight line A E makes with CD the two angles CE A, AED, these angles are together equal (I. 13) to two right angles. Again, because the straight line DE makes with A B the two angles AED, DEB, these angles are together equal (I. 13) to two right angles. But the two angles CEA, AED, have been proved to be equal to two right angles. Therefore the two angles CEA, AED, are equal (Ax. 1) to the two angles AED, DEB. From these equals, take away the common angle AED. Therefore the remaining angle CEA is equal (Ax. 3) to the remaining angle DE B. In the same manner, it can be demonstrated, that the angle CEB is equal to the angle AED. Therefore, if two straight lines, &c. Q. E. D. The corollaries added to this proposition by Euclid, are included in those now added to Prop. XIII. Excrcise 1.-If at a point in a straight line two other straight lines meet upon the opposite sides of it, and make the vertical or opposite angles equal, these two straight lines are in one and the same straight line. Exercise 2.-If two straight lines cut one another, and the vertical angles be bisected, the bisecting lines are in one and the same straight line. PROP. XVI. THEOREM. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. The Let A B C be a triangle, and let its side B C be produced to D. exterior angle ACD is greater than either of the interior opposite angles CBA, BAC. Bisect (I. 10) A C in E, join BE and produce it to F. Make E F equal (I. 3) to BE. Join FC. Because AE is equal (Const.) to EC, and BE (Const.) to E F. Therefore, in the triangles A E B, CEF, the the two sides A E, E B, are equal to the two sides CE, EF, each to each. But the angle A E B is equal (I. 15) to the angle CEF, because they are vertical angles. Therefore the base AB is equal (I. 4) to the base CF, the triangle AEB to the triangle CEF, and the remaining angles of the one to the remaining angles of the other, each to each:-viz., those to which |