Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/41

 Also, because ABD is a straight line, the angle DBE is equal to the angle EBA. Therefore the angle DBE is equal to the angle CBE, the less to the greater; which is impossible. Wherefore two straight lines cannot have a common segment. PROPOSITION 12. 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 the given point without it: it is required to draw from the point 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 the circle EGF, meeting AB at F and G. Bisect FG at H, 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, 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; therefore 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 the angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it. Wherefore a perpendicular CH has been drawn to the given straight line AB from the given point C without it.