PROPOSITION 10. PROBLEM. To bisect a given finite straight line, that is to divide it into two equal parts. Let AB be the given 1. straight line. It is required to divide it into two equal parts. Describe on it an equi lateral triangle ABC [I. 1], and bisect the angle ACB by the straight line CD. [I. 9.] AB' shall be cut into two equal A D parts at the point D. Because AC is equal to CB [Def. 24], and CD is common to the two triangles ACD, BCD, the two sides, AC, CD are equal to the two sides BC, CD, each to each ; and the angle ACD is equal to the angle BCD [Con.]; therefore the base AD is equal to tho base DB. [I. 4.] Therefore the straight line AB is divided into two equal parts at the point D. Q.E.F. Ex. The straight line which bisects the vertical angle of an isosceles triangle also bisects the base at right angles. 3 PROPOSITION 11. PROBLEM. To draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be 1. the given straight line, and C the given point in it. It is required to draw a straight line from the point Cat right angles to AB. А B Take any 2. . point D in AC, and make CE equal to CD. [1.3.] On DE describe the equilateral triangle DFE[I. 1), and join CF. The straight line CF drawn from the given point CĀ B shall be at right angles to the given straight line AB. 3. Because DC is equal to CE [Con.], and FC is 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 to the base EF (Def. 24.]; therefore the angle DCF is equal to the angle ECF [I. 8], 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 the angles is called a right angle [Def. 10]; therefore each of the angles DCF, ECF is a right angle. Therefore from the given point Cin the given straight line AB, CF has been drawn at right angles to AB. Q.E.F. If the point C were at the end of the line the construction could not be effected without producing the line. A method of drawing a line at right angles to a given line from its extremity and without producing the line will be pointed out in the note to prop. 32. Ex. Show how to find a point in a given line which shall be equi distant from two given points. (See Ex. to prop. 10.] 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 1. be produced to any length both ways, and let C be a given point without it. It is required to draw from C a straight line perpendicular to AB. Take any point D on the other side of AB, and from the centre C, at the distance CD, describe 2. ,E H A 3. the circle EGF, meeting AB in F and G. [Post. 3.] Bisect FG in H [I. 10], and join CH. The straight line CH drawn from the given point C shall be perpendicular to the given straight line AB. Join CF, CG. Because FH is equal to HG [Const.], and HC is common to the two triangles FHC, GHC; the two sides FH, HC are equal to the two sides GH, HC, each to each ; and the base CF is equal to the base CG [Def.]; therefore the angle CHF is equal to the angle CHG [I. 8.]; 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 the angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it. [Def. 10.] Therefore a perpendicular CH has been drawn to the given straight line AB from the given point C without it. Q.E.F. The given line must be of unlimited length to insure that the circle FDE will meet it in two points. If this restriction were removed the following construction might be adopted. In AB take any point D. Join DC, and with centre D and radius DC describe an arc CG. In AD take a point E ; join EC and describe an arc cutting the former arc in G. Join CG, which A shall be perpendicular to AB. Join EG and DG. The proof is ic'entical with exercise 2 to prop. 9. Ex. From a given point to draw a straight line equally inclined to two given straight lines which cut one another. [See Ex. to last two props.] PROPOSITION 19. THOREM. The angles which one straight line makes with another straight line on one side of it, either are two right angles, or are together equal to two right angles. Let the straight line AB make with the straight 1. line cd, on one side of it, the angles CBA, A E A ABD: these either are two right angles, or are together equal to two right angles. For if the angle CBA is equal to the angle ABD, each of them is a right angle. [Def. 10.] 2. But if not, from the point B draw BE at right angles to CD [I. ]; therefore, the angles CBE, EBD are two right angles. (Def. 10.] 3. Then because the angle CBE is equal to the two angles CBA, ABE; to each of these equals add the angle EBD; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [Ax. 2.] Again, the angle DBA is equal to the two angles DBE, EBA; to each of these equals add the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC. [Ax. 2.] But the angles CBE, EBD have been proved to be equal to the same three angles ; and things which are equal to the same thing are equal to one another [Ax. 1]; 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. Therefore, the angles, &c. Q.E.D. Ex. If the angles DBA and CBA in the figs. to this prop.) be bisected by straight lines BG and BH show that the angle GBH wil in each case be a right angle. If, at a point in a straight line, two other straight lines, on 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. |